Inequality Proof Problems [201-250]

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Restructured the IneqMath training data.

P201. Let $a, b, c \in \mathbb{R}^{+}$ such that $ab + bc + ca = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: $\frac{1}{a(a+b)}+\frac{1}{b(b+c)}+\frac{1}{c(c+a)} \geq C$

S201. $C = \frac{9}{2}$

The given inequality is equivalent to\n$\n\\frac{c(a+b)+a b}{a(a+b)}+\\frac{a(b+c)+b c}{b(b+c)}+\\frac{b(c+a)+a c}{c(c+a)} \\geq \\frac{9}{2}\n$\ni.e.\n$\n\\begin{align*}\n\\frac{a}{b} & +\\frac{b}{c}+\\frac{c}{a}+\\frac{b}{a+b}+\\frac{c}{b+c}+\\frac{a}{c+a} \\geq \\frac{9}{2} \\\\\n& \\Leftrightarrow \\quad \\frac{a+b}{b}+\\frac{b+c}{c}+\\frac{c+a}{a}+\\frac{b}{a+b}+\\frac{c}{b+c}+\\frac{a}{c+a} \\geq \\frac{15}{2} \\tag{1}\n\\end{align*}\n$\n\nWe have\n$\n\\begin{aligned}\n& \\frac{a+b}{b}+\\frac{b+c}{c}+\\frac{c+a}{a}+\\frac{b}{a+b}+\\frac{c}{b+c}+\\frac{a}{c+a} \\\\\n& \\quad=\\frac{a+b}{4 b}+\\frac{b+c}{4 c}+\\frac{c+a}{4 a}+\\frac{b}{a+b}+\\frac{c}{b+c}+\\frac{a}{c+a}\n\\end{aligned}\n$\n$\n\\begin{aligned}\n& +\\frac{3}{4}\\left(\\frac{a+b}{b}+\\frac{b+c}{c}+\\frac{c+a}{a}\\right) \\\\\n\\geq & 6 \\sqrt[6]{\\frac{a+b}{4 b} \\cdot \\frac{b+c}{4 c} \\cdot \\frac{c+a}{4 a} \\cdot \\frac{b}{a+b} \\cdot \\frac{c}{b+c} \\cdot \\frac{a}{c+a}}+\\frac{3}{4}\\left(\\frac{a}{b}+\\frac{b}{c}+\\frac{c}{a}+3\\right) \\\\\n\\geq & 3+\\frac{3}{4}\\left(3 \\sqrt[3]{\\frac{a}{b} \\cdot \\frac{b}{c} \\cdot \\frac{c}{a}}+3\\right)=3+\\frac{18}{4}=\\frac{15}{2}\n\\end{aligned}\n$\nas required.\n\nEquality holds when $a = b = c = \frac{1}{\sqrt{3}}$, since $ab + bc + ca = 3 \left(\frac{1}{\sqrt{3}}\right)^2 = 1$, and in this case, all the steps above become equalities. This gives the minimum value of the sum, so $C = \frac{9}{2}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{9}{2}$.

P202. Let $a, b, c$ be positive real numbers. Find the smallest positive number $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: $\frac{a+\sqrt{a b}+\sqrt[3]{a b c}}{C} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}$

S202. $C = 3$

Applying $A M \geq G M$ we get\n$\n\\sqrt[3]{a b \\cdot \\frac{a+b}{2}} \\geq \\sqrt[3]{a b \\cdot \\sqrt{a b}}=\\sqrt{a b}\n$\n\nSo\n$\na+\\sqrt{a b}+\\sqrt[3]{a b c} \\leq a+\\sqrt[3]{a b \\cdot \\frac{a+b}{2}}+\\sqrt[3]{a b c}\n$\n\nNow, it is enough to show that\n$\na+\\sqrt[3]{a b \\cdot \\frac{a+b}{2}}+\\sqrt[3]{a b c} \\leq 3 \\sqrt[3]{a \\cdot \\frac{a+b}{2} \\cdot \\frac{a+b+c}{3}}\n$\n\nAnother application of $A M \geq G M$ gives us\n$\sqrt[3]{1 \cdot \frac{2 a}{a+b} \cdot \frac{3 a}{a+b+c}} \leq \frac{1+\frac{2 a}{a+b}+\frac{3 a}{a+b+c}}{3}, \quad \sqrt[3]{1 \cdot 1 \cdot \frac{3 b}{a+b+c}} \leq \frac{2+\frac{3 b}{a+b+c}}{3}$ and\n$\n\\sqrt[3]{1 \\cdot \\frac{2 b}{a+b} \\cdot \\frac{3 c}{a+b+c}} \\leq \\frac{1+\\frac{2 b}{a+b}+\\frac{3 c}{a+b+c}}{3}\n$\n\nAdding, we obtain\n$\n\\sqrt[3]{\\frac{2 a}{a+b} \\cdot \\frac{3 a}{a+b+c}}+\\sqrt[3]{\\frac{3 b}{a+b+c}}+\\sqrt[3]{\\frac{2 b}{a+b} \\cdot \\frac{3 c}{a+b+c}} \\leq 3\n$\ni.e.\n$\n\\sqrt[3]{\\frac{1}{a} \\cdot \\frac{2}{a+b} \\cdot \\frac{3}{a+b+c}}\\left(a+\\sqrt[3]{a b \\cdot \\frac{a+b}{2}}+\\sqrt[3]{a b c}\\right) \\leq 3\n$\ni.e.\n$\na+\\sqrt[3]{a b \\cdot \\frac{a+b}{2}}+\\sqrt[3]{a b c} \\leq 3 \\sqrt[3]{a \\cdot \\frac{a+b}{2} \\cdot \\frac{a+b+c}{3}}\n$\n\nEquality holds when $a = b = c$, that is, when all variables are equal. In this case, both sides of the inequality are equal, and the minimum value of $C$ for which the inequality always holds is $C = 3$.\n\nTherefore, the answer is $C = 3$.

P203. Let $a, b, c, d, e$ be positive real numbers such that $a+b+c+d+e=5$. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c, d, e$ satisfying the given constraint: $a b c + b c d + c d e + d e a + e a b \leq C.$

S203. $C = 5$

Without loss of generality, we may assume that $e=\min \{a, b, c, d, e\}$.\nBy $AM \geq GM$, we have\n$\n\\begin{aligned}\na b c+b c d+c d e+d e a+e a b & =e(a+c)(b+d)+b c(a+d-e) \\\\\n& \\leq e\\left(\\frac{a+c+b+d}{2}\\right)^{2}+\\left(\\frac{b+c+a+d-e}{3}\\right)^{3} \\\\\n& =\\frac{e(5-e)^{2}}{4}+\\frac{(5-2 e)^{3}}{27}\n\\end{aligned}\n$\n\nSo it suffices to prove that\n$\n\\frac{e(5-e)^{2}}{4}+\\frac{(5-2 e)^{3}}{27} \\leq 5\n$\nwhich can be rewrite as $(e-1)^{2}(e+8) \geq 0$, which is obviously true.\nEquality holds if and only if $a = b = c = d = e = 1$, that is, when all variables are equal. In this case, $a b c + b c d + c d e + d e a + e a b = 5$. Thus, this gives the maximum value of the expression, so the minimal constant $C$ is $5$.\n\nTherefore, the answer is $C = 5$.

P204. Let $a, b, c$ be non-negative real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}_0$: $a^{3}+b^{3}+c^{3}+a b c \geq C(a+b+c)^{3}$

S204. $C = \frac{1}{7}$

We have\n$\n\\begin{aligned}\n(a+b+c)^{3} & =a^{3}+b^{3}+c^{3}+3\\left(a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\\right)+6 a b c \\\\\n& =\\frac{T[3,0,0]}{2}+3 T[2,1,0]+T[1,1,1]\n\\end{aligned}\n$\nand\n$\na^{3}+b^{3}+c^{3}+a b c=\\frac{T[3,0,0]}{2}+\\frac{T[1,1,1]}{6}\n$\n\nSo we need to prove that\n$\n7\\left(\\frac{T[3,0,0]}{2}+\\frac{T[1,1,1]}{6}\\right) \\geq \\frac{T[3,0,0]}{2}+3 T[2,1,0]+T[1,1,1]\n$\ni.e.\n$\n3 T[3,0,0]+\\frac{T[1,1,1]}{6} \\geq 3 T[2,1,0]\n$\nwhich is true according to $T[3,0,0] \geq T[2,1,0]$ and $T[1,1,1] \geq 0$ (Muirhead’s theorem).\n\nEquality holds when $a = b = c$, since then all symmetric sums are equal and the inequality becomes an equality. This gives the minimum value of $C$, so $C = \frac{1}{7}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{1}{7}$.

P205. Let $a, b, c, d > 1$ be real numbers. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c, d > 1$: $\sqrt{a-1}+\sqrt{b-1}+\sqrt{c-1}+\sqrt{d-1} \leq C \sqrt{(a b+1)(c d+1)}.$

S205. $C = 1$

We’ll prove that for every $x, y \in \mathbb{R}^{+}$ we have $\sqrt{x-1}+\sqrt{y-1} \leq \sqrt{x y}$.\nApplying the Cauchy-Schwarz inequality for $a_{1}=\sqrt{x-1}, a_{2}=1 ; b_{1}=1, b_{2}=\sqrt{y-1}$ gives us\n$\n(\\sqrt{x-1}+\\sqrt{y-1})^{2} \\leq x y, \\quad \\text { i.e. } \\quad \\sqrt{x-1}+\\sqrt{y-1} \\leq \\sqrt{x y}\n$\n\nNow we easily deduce that\n$\n\\sqrt{a-1}+\\sqrt{b-1}+\\sqrt{c-1}+\\sqrt{d-1} \\leq \\sqrt{a b}+\\sqrt{c d} \\leq \\sqrt{(a b+1)(c d+1)}\n$\n\nEquality holds when $a = b = c = d = 2$, since then $\sqrt{a-1} = 1$ and $\sqrt{a b + 1} = \sqrt{5}$, so both sides equal $4$ and $\sqrt{(2\cdot2+1)(2\cdot2+1)} = 5$. Thus, the minimum value of $C$ for which the inequality always holds is $C = 1$.\n\nTherefore, the answer is $C = 1$.

P206. Let $n > 2$ and let $x_{1}, x_{2}, \ldots, x_{n} > 0$ such that $\sum_{i=1}^{n} x_i = 1$. Find the maximal constant $C$ such that the following inequality holds for all $x_{i}$: $\prod_{i=1}^{n}\left(1+\frac{1}{x_{i}}\right) \geq C \prod_{i=1}^{n}\left(\frac{n-x_{i}}{1-x_{i}}\right).$

S206. $C =1$

The most natural idea is to use the fact that\n\n$\n\\frac{n-x_{i}}{1-x_{i}}=1+\\frac{n-1}{x_{1}+x_{2}+\\cdots+x_{i-1}+x_{i+1}+\\cdots+x_{n}}\n$\n\nThus, we have\n\n$\n\\prod_{i=1}^{n}\\left(\\frac{n-x_{i}}{1-x_{i}}\\right) \\leq \\prod_{i=1}^{n}\\left(1+\\frac{1}{\\sqrt[n-1]{x_{1} x_{2} \\ldots x_{i-1} x_{i+1} \\ldots x_{n}}}\\right)\n$\n\nand we have to prove the inequality\n\n$\n\\prod_{i=1}^{n}\\left(1+\\frac{1}{x_{i}}\\right) \\geq \\prod_{i=1}^{n}\\left(1+\\frac{1}{\\sqrt[n-1]{x_{1} x_{2} \\ldots x_{i-1} x_{i+1} \\ldots x_{n}}}\\right)\n$\n\nBut this one is not very hard, because it follows immediately by multiplying the inequalities\n\n$\n\\prod_{j \\neq i}\\left(1+\\frac{1}{x_{j}}\\right) \\geq\\left(1+\\sqrt[n-1]{\\prod_{j \\neq i} \\frac{1}{x_{j}}}\\right)^{n-1}\n$\n\nobtained from Huygens Inequality.,We will prove even more, that\n\n$\n\\prod_{i=1}^{n}\\left(1+\\frac{1}{x_{i}}\\right) \\geq\\left(\\frac{n^{2}-1}{n}\\right)^{n} \\cdot \\prod_{i=1}^{n} \\frac{1}{1-x_{i}}\n$\n\nIt is clear that this inequality is stronger than the initial one. First, let us prove that\n\n$\n\\prod_{i=1}^{n} \\frac{1+x_{i}}{1-x_{i}} \\geq\\left(\\frac{n+1}{n-1}\\right)^{n}\n$\n\nThis follows from Jensen’s Inequality for the convex function $f(x)=\ln (1+x)-$ $\ln (1-x)$. So, it suffices to prove that\n\n$\n\\frac{\\left(\\frac{n+1}{n-1}\\right)^{n}}{\\prod_{i=1}^{n} x_{i}} \\cdot \\prod_{i=1}^{n}\\left(1-x_{i}\\right)^{2} \\geq\\left(\\frac{n^{2}-1}{n}\\right)^{n}\n$\n\nBut a quick look shows that this is exactly the inequality proved in the solution of the problem 121.\n\nEquality holds when $x_1 = x_2 = \cdots = x_n = \frac{1}{n}$, since then all terms are equal and the AM-GM and Jensen inequalities used above become equalities. This gives the minimum value of $C$ for which the inequality always holds, so the maximal constant is $C = 1$.\n\nTherefore, the answer is $C = 1$.

P207. Let $a, b, c \geq 0$ such that $a + b + c = 1$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$: $a^{2} + b^{2} + c^{2} \leq C(a^{3} + b^{3} + c^{3}) + 3abc$

S207. $C = 2$

Let $p=a+b+c=1, q=a b+b c+c a, r=a b c$.\nThe given inequality is equivalent to\n$\n1-2 q \\leq 2(1-3 q+3 r)+3 r \\quad \\Leftrightarrow \\quad 4 q \\leq 1+9 r\n$\n\nEquality holds when two of the variables are zero and the third is $1$, i.e., $(a, b, c) = (1, 0, 0)$ and its permutations. In this case, $q = 0$ and $r = 0$, so $4q = 0$ and $1 + 9r = 1$, and the inequality becomes $0 \leq 1$, which is tight. This gives the minimum value of $C$.\n\nTherefore, the answer is $C = 2$.

P208. Let $a, b, c, d > 0$ be real numbers such that $a \leq 1$, $a+b \leq 5$, $a+b+c \leq 14$, and $a+b+c+d \leq 30$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c, d$ satisfying the given constraints: $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} \leq C$

S208. $C = 10$

The function $f:(0,+\infty) \rightarrow(0,+\infty)$ defined by $f(x)=\sqrt{x}$ is concave on $(0,+\infty)$, so by Jensen’s inequality, for\n$\nn=4, \\quad \\alpha_{1}=\\frac{1}{10}, \\quad \\alpha_{2}=\\frac{2}{10}, \\quad \\alpha_{3}=\\frac{3}{10}, \\quad \\alpha_{4}=\\frac{4}{10}\n$\nwe get\n$\n\\frac{1}{10} \\sqrt{a}+\\frac{2}{10} \\sqrt{\\frac{b}{4}}+\\frac{3}{10} \\sqrt{\\frac{c}{9}}+\\frac{4}{10} \\sqrt{\\frac{d}{16}} \\leq \\sqrt{\\frac{a}{10}+\\frac{b}{20}+\\frac{c}{30}+\\frac{d}{40}}\n$\ni.e.\n$\n\\begin{equation*}\n\\sqrt{a}+\\sqrt{b}+\\sqrt{c}+\\sqrt{d} \\leq 10 \\sqrt{\\frac{12 a+6 b+4 c+3 d}{120}} \\tag{1}\n\\end{equation*}\n$\n\nOn the other hand, we have\n$\n\\begin{aligned}\n& 12 a+6 b+4 c+3 d \\\\\n& \\quad=3(a+b+c+d)+(a+b+c)+2(a+b)+6 a \\\\\n& \\quad \\leq 3 \\cdot 30+14+2 \\cdot 5+6 \\cdot 1=120\n\\end{aligned}\n$\n\nBy (1) and the last inequality we obtain the required result.\n\nEquality holds when $a = 1$, $b = 4$, $c = 9$, $d = 16$, since these values satisfy all the constraints with equality and yield $\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{d} = 1+2+3+4 = 10$. This gives the maximum value of the sum, so $C = 10$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 10$.

P209. Prove that if $n \geq 2$ and $a_{1}, a_{2}, \ldots, a_{n}$ are real numbers with product 1, then find the largest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$: $a_{1}^{2}+a_{2}^{2}+\cdots+a_{n}^{2}-n \geq C \cdot\frac{n}{n-1} \cdot \sqrt[n]{n-1}\left(a_{1}+a_{2}+\cdots+a_{n}-n\right)$

S209. $C = 2$

We will prove the inequality by induction. For $n=2$ it is trivial. Now, suppose the inequality is true for $n-1$ numbers and let us prove it for $n$. First, it is easy to see that it is enough to prove it for $a_{1}, \ldots, a_{n}>0$ (otherwise we replace $a_{1}, a_{2}, \ldots, a_{n}$ with $ㅣa_{1}ㅣ,ㅣa_{2}ㅣ, \ldots,ㅣa_{n}ㅣ$, which have product 1 . Yet, the right hand side increases). Now, let $a_{n}$ the maximum number among $a_{1}, a_{2}, \ldots, a_{n}$ and let $G$ the geometric mean of $a_{1}, a_{2}, \ldots, a_{n-1}$. First, we will prove that\n\n$\n\\begin{gathered}\na_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{1}+a_{2}+\\cdots+a_{n}-n\\right) \\geq \\\\\n\\geq a_{n}^{2}+(n-1) G^{2}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{n}+(n-1) G-n\\right)\n\\end{gathered}\n$\n\nwhich is equivalent to\n\n$\na_{1}^{2}+a_{2}^{2}+\\cdots+a_{n-1}^{2}-(n-1) \\sqrt[n-1]{a_{1}^{2} a_{2}^{2} \\ldots a_{n-1}^{2}} \\geq\n$\n\n$\n\\geq \\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{1}+a_{2}+\\cdots+a_{n-1}-(n-1) \\sqrt[n-1]{a_{1} a_{2} \\ldots a_{n-1}}\\right)\n$\n\nBecause, $\sqrt[n-1]{a_{1} a_{2} \ldots a_{n-1}} \leq 1$ and $a_{1}+a_{2}+\cdots+a_{n-1}-(n-1) \sqrt[n-1]{a_{1} a_{2} \ldots a_{n-1}} \geq$ 0 , it is enough to prove the inequality\n\n$\na_{1}^{2}+\\cdots+a_{n-1}^{2}-(n-1) G^{2} \\geq \\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1} \\cdot G \\cdot\\left(a_{1}+\\cdots+a_{n-1}-(n-1) G\\right)\n$\n\nNow, we apply the inductive hypothesis for the numbers $\frac{a_{1}}{G}, \ldots, \frac{a_{n-1}}{G}$ which have product 1 and we infer that\n\n$\n\\frac{a_{1}^{2}+\\cdots+a_{n-1}^{2}}{G^{2}}-n+1 \\geq \\frac{2(n-1)}{n-2} \\cdot \\sqrt[n-1]{n-2}\\left(\\frac{a_{1}+\\cdots+a_{n-1}}{G}-n+1\\right)\n$\n\nand so it suffices to prove that\n\n$\n\\frac{2(n-1)}{n-2} \\cdot \\sqrt[n-1]{n-2}\\left(a_{1}+\\cdots+a_{n-1}-(n-1) G\\right) \\geq \\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{1}+\\cdots+a_{n-1}-(n-1) G\\right)\n$\n\nwhich is the same as $1+\frac{1}{n(n-2)} \geq \frac{\sqrt[n]{n-1}}{\sqrt[n-1]{n-2}}$. This becomes\n\n$\n\\left(1+\\frac{1}{n(n-2)}\\right)^{n(n-1)} \\geq \\frac{(n-1)^{n-1}}{(n-2)^{n}}\n$\n\nand it follows for $n>4$ from\n\n$\n\\left(1+\\frac{1}{n(n-2)}\\right)^{n(n-1)}>2\n$\n\nand\n\n$\n\\frac{(n-1)^{n-1}}{(n-2)^{n}}=\\frac{1}{n-2} \\cdot\\left(1+\\frac{1}{n-2}\\right) \\cdot\\left(1+\\frac{1}{n-2}\\right)^{n-2}<\\frac{e}{n-2}\\left(1+\\frac{1}{n-2}\\right)<2\n$\n\nFor $n=3$ and $n=4$ it is easy to check.\nThus, we have proved that\n\n$\n\\begin{aligned}\na_{1}^{2} & +a_{2}^{2}+\\cdots+a_{n}^{2}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{1}+a_{2}+\\cdots+a_{n}-n\\right) \\geq \\\\\n& \\geq a_{n}^{2}+(n-1) G^{2}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(a_{n}+(n-1) G-n\\right)\n\\end{aligned}\n$\n\nand it is enough to prove that\n\n$\nx^{2(n-1)}+\\frac{n-1}{x^{2}}-n \\geq \\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(x^{n-1}+\\frac{n-1}{x}-n\\right)\n$\n\nfor all $x \geq 1$ (we took $x=\frac{1}{G}$ ). Let us consider the function\n\n$\nf(x)=x^{2(n-1)}+\\frac{n-1}{x^{2}}-n-\\frac{2 n}{n-1} \\cdot \\sqrt[n]{n-1}\\left(x^{n-1}+\\frac{n-1}{x}-n\\right)\n$\n\nWe have\n\n$\nf^{\\prime}(x)=2 \\cdot \\frac{x^{n}-1}{x^{2}} \\cdot\\left[\\frac{(n-1)\\left(x^{n}+1\\right)}{x}-n \\sqrt[n]{n-1}\\right] \\geq 0\n$\n\nbecause\n\n$\nx^{n-1}+\\frac{1}{x}=x^{n-1}+\\frac{1}{(n-1) x}+\\cdots+\\frac{1}{(n-1) x} \\geq n \\sqrt[n]{\\frac{1}{(n-1)^{n-1}}}\n$\n\nThus, $f$ is increasing and so $f(x) \geq f(1)=0$. This proves the inequality.\n\nEquality holds when $a_1 = a_2 = \cdots = a_n = 1$, since then the product is 1, $a_1^2 + \cdots + a_n^2 = n$, and $a_1 + \cdots + a_n = n$, so both sides of the inequality are zero. This gives the minimum value of the left side, so $C = 2$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

P210. Let $a, b, c$ be positive real numbers. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: $\frac{a^{3}}{b^{2}}+\frac{b^{3}}{c^{2}}+\frac{c^{3}}{a^{2}} \geq C \left( \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a} \right)$

S210. $C = 1$

By the Cauchy-Schwarz Inequality we have\n\n$\n(a+b+c)\\left(\\frac{a^{3}}{b^{2}}+\\frac{b^{3}}{c^{2}}+\\frac{c^{3}}{a^{2}}\\right) \\geq\\left(\\frac{a^{2}}{b}+\\frac{b^{2}}{c}+\\frac{c^{2}}{a}\\right)^{2}\n$\n\nso we only have to prove that\n\n$\n\\frac{a^{2}}{b}+\\frac{b^{2}}{c}+\\frac{c^{2}}{a} \\geq a+b+c\n$\n\nBut this follows immediately from the Cauchy-Schwarz Inequality.\n\nEquality holds when $a = b = c$, in which case both sides of the original inequality are equal. This gives the minimum value of the left side relative to the right, so $C = 1$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$. “, “We have\n\n$\n\\frac{a^{3}}{b^{2}} \\geq \\frac{a^{2}}{b}+a-b \\Leftrightarrow a^{3}+b^{3} \\geq a b(a+b) \\Leftrightarrow(a-b)^{2}(a+b) \\geq 0\n$\n\nwhich is clearly true. Writing the analogous inequalities and adding them up gives\n\n$\n\\frac{a^{3}}{b^{2}}+\\frac{b^{3}}{c^{2}}+\\frac{c^{3}}{a^{2}} \\geq \\frac{a^{2}}{b}+a-b+\\frac{b^{2}}{c}+b-c+\\frac{c^{2}}{a}+c-a=\\frac{a^{2}}{b}+\\frac{b^{2}}{c}+\\frac{c^{2}}{a}\n$\n\nEquality holds when $a = b = c$, since then both sides are equal. This gives the minimum value of the left side relative to the right, so the maximal constant is $C = 1$.\n\nTherefore, the answer is $C = 1$.

P211. Let $a, b, c$ be positive real numbers such that $a b c \geq 1$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$: $a+b+c \geq C\left(\frac{1+a}{1+b}+\frac{1+b}{1+c}+\frac{1+c}{1+a}\right).$

S211. $C = 1$

We have\n$\n\\begin{aligned}\na & +b+c-\\frac{1+a}{1+b}-\\frac{1+b}{1+c}-\\frac{1+c}{1+a} \\\\\n& =(1+a)\\left(1-\\frac{1}{1+b}\\right)+(1+b)\\left(1-\\frac{1}{1+c}\\right)+(1+c)\\left(1-\\frac{1}{1+a}\\right)-3 \\\\\n& =\\frac{(1+a) b}{1+b}+\\frac{(1+b) c}{1+c}+\\frac{(1+c) a}{1+a}-3 \\\\\n& \\geq 3 \\sqrt[3]{\\frac{(1+a) b}{1+b} \\cdot \\frac{(1+b) c}{1+c} \\cdot \\frac{(1+c) a}{1+a}}-3 \\\\\n& =3 \\sqrt[3]{a b c}-3 \\geq 0 \\quad(a b c \\geq 1)\n\\end{aligned}\n$\n\nEquality holds when $a = b = c = 1$, since then $a + b + c = 3$ and $\frac{1+a}{1+b} + \frac{1+b}{1+c} + \frac{1+c}{1+a} = 3$, so the inequality becomes $3 \geq C \cdot 3$, which gives $C = 1$. This is the maximum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P212. Let $h_{a}, h_{b}$, and $h_{c}$ be the lengths of the altitudes, and $R$ and $r$ be the circumradius and inradius, respectively, of a given triangle. Determine the minimal constant $C$ such that the following inequality holds for all triangles: $h_{a}+h_{b}+h_{c} \leq 2R+Cr.$

S212. $C = 5$

Lemma 21.4 In an arbitrary triangle we have\n$\na b+b c+c a=r^{2}+s^{2}+4 r R \\quad \\text { and } \\quad a^{2}+b^{2}+c^{2}=2\\left(s^{2}-4 R r-r^{2}\\right)\n$\n\nProof We have\n$\n\\begin{aligned}\nr^{2}+s^{2}+4 r R & =\\frac{P^{2}}{s^{2}}+s^{2}+\\frac{a b c}{P} \\cdot \\frac{P}{s}=\\frac{(s-a)(s-b)(s-c)}{s}+s^{2}+\\frac{a b c}{s} \\\\\n& =\\frac{s^{3}-a s^{2}-b s^{2}-c s^{2}+a b s+b c s+c a s-a b c+s^{3}+a b c}{s} \\\\\n& =2 s^{2}-s(a+b+c)+a b+b c+c a \\\\\n& =2 s^{2}-2 s^{2}+a b+b c+c a=a b+b c+c a\n\\end{aligned}\n$\n\nHence\n$\n\\begin{equation*}\na b+b c+c a=r^{2}+s^{2}+4 r R \\tag{1}\n\\end{equation*}\n$\n\nNow by (1) we have\n$\n\\begin{aligned}\na b+b c+c a & =r^{2}+s^{2}+4 r R=\\frac{1}{2}\\left(2 r^{2}+8 r R+\\frac{(a+b+c)^{2}}{2}\\right) \\\\\n& =\\frac{1}{2}\\left(2 r^{2}+8 r R+\\frac{a^{2}+b^{2}+c^{2}}{2}\\right)+\\frac{a b+b c+c a}{2}\n\\end{aligned}\n$\nfrom which it follows that\n$\n\\begin{equation*}\na b+b c+c a=2 r^{2}+8 r R+\\frac{a^{2}+b^{2}+c^{2}}{2} \\tag{2}\n\\end{equation*}\n$\n\nNow (1) and (2) yields\n$\n\\begin{equation*}\na^{2}+b^{2}+c^{2}=2\\left(s^{2}-4 R r-r^{2}\\right) \\tag{3}\n\\end{equation*}\n$\n\nWithout proof we will give the following lemma.\n\nIn an arbitrary triangle we have\n$\n\\begin{equation*}\ns^{2} \\leq 4 R^{2}+4 R r+3 r^{2} \\tag{4}\n\\end{equation*}\n$\n\nIn an arbitrary triangle we have $a^{2}+b^{2}+c^{2} \leq 8 R^{2}+4 r^{2}$.\nProof From (3) and (4) we have\n$\na^{2}+b^{2}+c^{2}=2\\left(s^{2}-4 R r-r^{2}\\right) \\leq 2\\left(4 R^{2}+4 R r+3 r^{2}-4 R r-r^{2}\\right)=8 R^{2}+4 r^{2}\n$\n\nHence\n$\n\\begin{equation*}\na^{2}+b^{2}+c^{2} \\leq 8 R^{2}+4 r^{2} \\tag{5}\n\\end{equation*}\n$\n\nNow let us consider our problem.\nWe have\n$\n\\begin{aligned}\n& 2 R\\left(h_{a}+h_{b}+h_{c}\\right)=2 R\\left(\\frac{2 P}{a}+\\frac{2 P}{b}+\\frac{2 P}{c}\\right)=4 P R \\frac{a b+b c+c a}{a b c} \\\\\n&=a b+b c+c a \\\\\n& \\stackrel{(2)}{=} 2 r^{2}+8 r R+\\frac{a^{2}+b^{2}+c^{2}}{2} \\\\\n& \\stackrel{(4)}{\\leq} 2 r^{2}+8 r R+4 R^{2}+2 r^{2} \\\\\n& \\Leftrightarrow \\quad R\\left(h_{a}+h_{b}+h_{c}\\right) \\leq 2 R^{2}+4 R r+2 r^{2} \\leq 2 R^{2}+4 R r+R r \\leq R(2 R+5 r)\n\\end{aligned}\n$\n\nHence\n$\nh_{a}+h_{b}+h_{c} \\leq 2 R+5 r\n$\n\nEquality occurs iff $a=b=c$.\n\nEquality holds when the triangle is equilateral, i.e., $a = b = c$. In this case, the bound is achieved, and $C = 5$ gives the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 5$.

P213. Let $a, b, c > 0$ be real numbers such that $a + b + c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: $\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b} \geq C$

S213. $C = 2$

Applying the Cauchy-Schwarz inequality for the sequences\n$\na_{1}=\\sqrt{\\frac{a^{2}+b}{b+c}}, \\quad a_{2}=\\sqrt{\\frac{b^{2}+c}{c+a}}, \\quad a_{3}=\\sqrt{\\frac{c^{2}+a}{a+b}}\n$\nand\n$b_{1}=\sqrt{\left(a^{2}+b\right)(b+c)}, \quad b_{2}=\sqrt{\left(b^{2}+c\right)(c+a)}, \quad b_{3}=\sqrt{\left(c^{2}+a\right)(a+b)}$\nwe obtain\n$\frac{a^{2}+b}{b+c}+\frac{b^{2}+c}{c+a}+\frac{c^{2}+a}{a+b} \geq \frac{\left(a^{2}+b^{2}+c^{2}+1\right)^{2}}{\left(a^{2}+b\right)(b+c)+\left(b^{2}+c\right)(c+a)+\left(c^{2}+a\right)(a+b)}$.\nSo it suffices to show that\n$\n\\frac{\\left(a^{2}+b^{2}+c^{2}+1\\right)^{2}}{\\left(a^{2}+b\\right)(b+c)+\\left(b^{2}+c\\right)(c+a)+\\left(c^{2}+a\\right)(a+b)} \\geq 2\n$\n\nWe have\n$\n\\begin{aligned}\n& \\frac{\\left(a^{2}+b^{2}+c^{2}\\right.}{\\left(a^{2}+b\\right)(b+c)+\\left(b^{2}+c\\right)(c+a)+\\left(c^{2}+a\\right)(a+b)} \\geq 2 \\\\\n& \\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}+1\\right)^{2} \\geq 2\\left(\\left(a^{2}+b\\right)(b+c)+\\left(b^{2}+c\\right)(c+a)\\right. \\\\\n&\\left.+\\left(c^{2}+a\\right)(a+b)\\right) \\\\\n& \\Leftrightarrow \\quad 1+\\left(a^{2}+b^{2}+c^{2}\\right)^{2} \\geq 2\\left(a^{2}(b+c)+b^{2}(c+a)+c^{2}(a+b)\\right) \\\\\n&+2(a b+b c+c a) \\\\\n& \\Leftrightarrow \\quad 1+\\left(a^{2}+b^{2}+c^{2}\\right)^{2} \\geq 2\\left(a^{2}(1-a)+b^{2}(1-b)+c^{2}(1-c)\\right) \\\\\n&+2(a b+b c+c a)\n\\end{aligned}\n$\n$\n\\begin{aligned}\n\\Leftrightarrow \\quad 1+\\left(a^{2}+b^{2}+c^{2}\\right)^{2} \\geq & 2\\left(a^{2}+b^{2}+c^{2}-a^{3}-b^{3}-c^{3}\\right) \\\\\n& +2(a b+b c+c a) \\\\\n\\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq & 2\\left(a^{2}+b^{2}+c^{2}+a b+b c+c a\\right)-1 \\\\\n\\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq & 2(a(1-c)+b(1-a) \\\\\n& +c(1-b))-1 \\\\\n\\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq & 1-2(a b+b c+c a) \\\\\n\\Leftrightarrow \\quad\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq & \\geq(a+b+c)^{2}-2(a b+b c+c a) \\\\\n& =a^{2}+b^{2}+c^{2} .\n\\end{aligned}\n$\n\nSo we need to show that\n$\n\\begin{equation*}\n\\left(a^{2}+b^{2}+c^{2}\\right)^{2}+2\\left(a^{3}+b^{3}+c^{3}\\right) \\geq a^{2}+b^{2}+c^{2} \\tag{1}\n\\end{equation*}\n$\n\nBy Chebishev’s inequality we deduce\n$(a+b+c)\left(a^{2}+b^{2}+c^{2}\right) \leq 3\left(a^{3}+b^{3}+c^{3}\right), \quad$ i.e. $\quad a^{3}+b^{3}+c^{3} \geq \frac{a^{2}+b^{2}+c^{2}}{3}$,\nand clearly $\left(a^{2}+b^{2}+c^{2}\right)^{2} \geq \frac{a^{2}+b^{2}+c^{2}}{3}$.\nAdding these inequalities gives us inequality (1).2 Take $a+b+c=p=1, a b+b c+c a=q, a b c=r$ and use the method from Chap. 14.\n\nEquality holds when $a = b = c = \frac{1}{3}$, since then $a + b + c = 1$ and the expression evaluates to $2$. This gives the minimum value of the sum, so $C = 2$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

P214. Let $a_{1}, a_{2}, \ldots, a_{n}>1$ be positive integers. Find the minimal constant $C$ such that at least one of the numbers $\sqrt[a_{1}]{a_{2}}, \sqrt[a_{2}]{a_{3}}, \ldots, \sqrt[a_{n}]{\sqrt{a_{n}}}, \sqrt[a_{n}]{a_{1}}$ is less than or equal to $C$ for all sequences of positive integers $a_1, a_2, \ldots, a_n > 1$: $\min \left( \sqrt[a_{1}]{a_{2}}, \sqrt[a_{2}]{a_{3}}, \ldots, \sqrt[a_{n}]{\sqrt{a_{n}}}, \sqrt[a_{n}]{a_{1}} \right) \leq C$

S214. $C = \sqrt[3]{3}$

Suppose we have $a_{i+1}^{\frac{1}{a_{i}}}>3^{\frac{1}{3}}$ for all $i$. First, we will prove that $n^{\frac{1}{n}} \leq 3^{\frac{1}{3}}$ for all natural number $n$. For $n=1,2,3,4$ it is clear. Suppose the inequality is true for $n>3$ and let us prove it for $n+1$. This follows from the fact that\n\n$\n1+\\frac{1}{n} \\leq 1+\\frac{1}{4}<\\sqrt[3]{3} \\Rightarrow 3^{\\frac{n+1}{3}}=\\sqrt[3]{3} \\cdot 3^{\\frac{n}{3}} \\geq \\frac{n+1}{n} \\cdot n=n+1\n$\n\nThus, using this observation, we find that $a_{i+1}^{\frac{1}{a_{i}}}>3^{\frac{1}{3}} \geq a_{i+1}^{\frac{1}{a_{i+1}}} \Rightarrow a_{i+1}>a_{i}$ for all $i$, which means that $a_{1}<a_{2}<\cdots<a_{n-1}<a_{n}<a_{1}$, contradiction.\n\nEquality holds when $a_i = 3$ and $a_{i+1} = 3$ for some $i$, so $\sqrt[3]{3} = 3^{1/3}$. This gives the minimal possible value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt[3]{3}$.

P215. Let $a, b, c, d$ be positive real numbers such that $a^2 + b^2 + c^2 + d^2 = 4$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c, d$: $\frac{a^2 + b^2 + 3}{a + b} + \frac{b^2 + c^2 + 3}{b + c} + \frac{c^2 + d^2 + 3}{c + d} + \frac{d^2 + a^2 + 3}{d + a} \geq C.$

S215. $C = 10$

Observe that for any real numbers $x, y$ we have\n$\nx^{2}+x y+y^{2}=\\left(x+\\frac{y}{2}\\right)^{2}+\\frac{3 y^{2}}{4} \\geq 0\n$\nequality achieves if and only if $x=y=0$.\nHence $(a-1)^{2}+(a-1)(b-1)+(b-1)^{2} \geq 0$, which is equivalent to\n$\na^{2}+b^{2}+a b-3 a-3 b+3 \\geq 0\n$\nfrom which we obtain\n$\na^{2}+b^{2}+3 \\geq 3 a+3 b-a b\n$\ni.e.\n$\n\\frac{a^{2}+b^{2}+3}{a+b} \\geq 3-\\frac{a b}{a+b}\n$\n\nBy $A M \geq G M$ we easily deduce that\n$\n\\frac{a+b}{4} \\geq \\frac{a b}{a+b}\n$\n\nTherefore by previous inequality we get\n$\n\\frac{a^{2}+b^{2}+3}{a+b} \\geq 3-\\frac{a+b}{4}\n$\n\nSimilarly we obtain\n$\n\\begin{aligned}\n& \\frac{b^{2}+c^{2}+3}{b+c} \\geq 3-\\frac{b+c}{4}, \\quad \\frac{c^{2}+d^{2}+3}{c+d} \\geq 3-\\frac{c+d}{4} \\quad \\text { and } \\\\\n& \\frac{d^{2}+a^{2}+3}{d+a} \\geq 3-\\frac{d+a}{4}\n\\end{aligned}\n$\n\nAdding the last four inequality yields\n$\n\\begin{equation*}\n\\frac{a^{2}+b^{2}+3}{a+b}+\\frac{b^{2}+c^{2}+3}{b+c}+\\frac{c^{2}+d^{2}+3}{c+d}+\\frac{d^{2}+a^{2}+3}{d+a} \\geq 12-\\frac{a+b+c+d}{2} \\tag{1}\n\\end{equation*}\n$\n\nAccording to inequality $Q M \geq A M$ we deduce that\n$\n\\sqrt{\\frac{a^{2}+b^{2}+c^{2}+d^{2}}{4}} \\geq \\frac{a+b+c+d}{4}\n$\nand since $a^{2}+b^{2}+c^{2}+d^{2}=4$ we obtain\n$\n\\begin{equation*}\na+b+c+d \\leq 4 \\tag{2}\n\\end{equation*}\n$\n\nBy (1) and (2) we get\n$\n\\begin{aligned}\n\\frac{a^{2}+b^{2}+3}{a+b}+\\frac{b^{2}+c^{2}+3}{b+c}+\\frac{c^{2}+d^{2}+3}{c+d}+\\frac{d^{2}+a^{2}+3}{d+a} & \\geq 12-\\frac{a+b+c+d}{2} \\\\\n& \\geq 12-\\frac{4}{2}=10\n\\end{aligned}\n$\nas required.\n\nEquality occurs if and only if $a = b = c = d = 1$, since then $a^2 + b^2 + c^2 + d^2 = 4$ and $a + b + c + d = 4$, so the minimum value of the sum is achieved and $C = 10$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 10$.

P216. Let $a, b, c, d, e$ be non-negative real numbers such that $a + b + c + d + e = 5$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c, d, e$ satisfying the given constraint: $4\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\right)+5abcd \geq C$

S216. $C = 25$

Without loss of generality we may assume that $a \geq b \geq c \geq d \geq e$.\nLet us denote\n$\nf(a, b, c, d, e)=4\\left(a^{2}+b^{2}+c^{2}+d^{2}+e^{2}\\right)+5 a b c d\n$\n\nThen we easily deduce that\n$\n\\begin{equation*}\nf(a, b, c, d, e)-f\\left(\\frac{a+d}{2}, b, c, \\frac{a+d}{2}, e\\right)=\\frac{(a-d)^{2}}{4}(8-5 b c e) \\tag{1}\n\\end{equation*}\n$\n\nSince $a \geq b \geq c \geq d \geq e$, we have\n$\n3 \\sqrt[3]{b c e} \\leq b+c+e \\leq \\frac{3(a+b+c+d+e)}{5}=3\n$\n\nThus it follows that bce $\leq 1$.\nNow, by (1) and the last inequality we get\n$\n\\begin{aligned}\nf(a, b, c, d, e)-f\\left(\\frac{a+d}{2}, b, c, \\frac{a+d}{2}, e\\right) & =\\frac{(a-d)^{2}}{4}(8-5 b c e) \\\\\n& \\geq \\frac{(a-d)^{2}}{4}(8-5) \\geq 0\n\\end{aligned}\n$\ni.e.\n$\nf(a, b, c, d, e) \\geq f\\left(\\frac{a+d}{2}, b, c, \\frac{a+d}{2}, e\\right)\n$\n\nAccording to the SMV theorem it remains to prove that $f(t, t, t, t, e) \geq 25$, under the condition $4 t+e=5$.\n\nClearly $4 t \leq 5$.\n\nWe have\n$\n\\begin{aligned}\n& f(t, t, t, t, e) \\geq 25 \\\\\n& \\quad \\Leftrightarrow \\quad 4\\left(t^{2}+e^{2}\\right)+5 t^{4} e \\geq 25 \\\\\n& \\quad \\Leftrightarrow \\quad 4 t^{2}+4(5-4 t)^{2}+5 t^{4}(5-4 t)-25 \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad(5-4 t)(t-1)^{2}\\left(t^{2}+2 t+3\\right) \\geq 0\n\\end{aligned}\n$\nwhich is true.\nEquality occurs if and only if $a=b=c=d=e=1$ or $a=b=c=d=5/4,\ e=0$ (up to permutation).\n\nEquality holds when either all variables are equal, i.e., $a = b = c = d = e = 1$, or when four variables are $5/4$ and the fifth is $0$, i.e., $a = b = c = d = 5/4$, $e = 0$ (up to permutation). In both cases, the expression attains its minimum value, so $C = 25$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 25$.

P217. Let $a, b, c$ be positive real numbers. Find the minimal constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: $a^{b} b^{c} c^{a} \leq\left(C(a+b+c)\right)^{a+b+c}$

S217. $C = \frac{1}{3}$

By the weighted power mean inequality we have\n$\n\\begin{aligned}\n\\left(a^{b} b^{c} c^{a}\\right)^{\\frac{1}{a+b+c}} & =a^{\\frac{b}{a+b+c}} \\cdot b^{\\frac{c}{a+b+c}} \\cdot c^{\\frac{a}{a+b+c}} \\leq \\frac{b a+c b+a c}{a+b+c} \\leq \\frac{(a+b+c)^{2}}{3(a+b+c)} \\\\\n& =\\frac{a+b+c}{3}\n\\end{aligned}\n$\n\nEquality holds when $a = b = c$, in which case $a^b b^c c^a = a^{a} a^{a} a^{a} = a^{3a}$ and $(C(a+b+c))^{a+b+c} = (3C a)^{3a}$, so $a^{3a} \leq (3C a)^{3a}$, which gives $C \geq \frac{1}{3}$. Thus, the minimal value of $C$ is $\frac{1}{3}$.\n\nTherefore, the answer is $C = \frac{1}{3}$.

P218. Let $a, b, c$ be the lengths of the sides of a triangle. Find the constant $C$ such that the following equation holds for all $a, b, c$: $2(a b^{2} + b c^{2} + c a^{2}) = a^{2} b + b^{2} c + c^{2} a + C a b c$ and ensures that the triangle is equilateral.

S218. $C = 3$

We’ll show that\n$\na^{2} b+b^{2} c+c^{2} a+3 a b c \\geq 2\\left(a b^{2}+b c^{2}+c a^{2}\\right)\n$\nwith equality if and only if $a=b=c$, i.e. the triangle is equilateral.\nLet us use Ravi’s substitutions, i.e. $a=x+y, b=y+z, c=z+x$. Then the given inequality becomes\n$\nx^{3}+y^{3}+z^{3}+x^{2} y+y^{2} z+z^{2} x \\geq 2\\left(x^{2} z+y^{2} x+z^{2} y\\right)\n$\n\nSince $A M \geq G M$ we have\n$\nx^{3}+z^{2} x \\geq 2 x^{2} z,\\quad y^{3}+x^{2} y \\geq 2 y^{2} x,\\quad z^{3}+y^{2} z \\geq 2 z^{2} y\n$\n\nAfter adding these inequalities we obtain\n$\nx^{3}+y^{3}+z^{3}+x^{2} y+y^{2} z+z^{2} x \\geq 2\\left(x^{2} z+y^{2} x+z^{2} y\\right)\n$\n\nEquality holds if and only if $x = y = z$, which means $a = b = c$, i.e., the triangle is equilateral. This gives the unique value $C = 3$ for which the equation holds for all $a, b, c$ and ensures the triangle is equilateral.\n\nTherefore, the answer is $C = 3$.

P219. Let $D, E$ and $F$ be the feet of the altitudes of the triangle $ABC$ dropped from the vertices $A, B$ and $C$, respectively. Determine the largest constant $C$ such that the following inequality holds for all triangles $ABC$: $\left(\frac{\overline{EF}}{a}\right)^{2}+\left(\frac{\overline{FD}}{b}\right)^{2}+\left(\frac{\overline{DE}}{c}\right)^{2} \geq C$

S219. $C = \frac{3}{4}$

Clearly $\overline{E F}=a \cos \alpha, \overline{F D}=b \cos \beta, \overline{D E}=c \cos \gamma$, and the given inequality becomes\n$\n\\cos ^{2} \\alpha+\\cos ^{2} \\beta+\\cos ^{2} \\gamma \\geq \\frac{3}{4}\n$\n\nEquality holds when the triangle is equilateral, i.e., $\alpha = \beta = \gamma = 60^\circ$, so $\cos^2 60^\circ = (1/2)^2 = 1/4$ and the sum is $3 \times 1/4 = 3/4$. This gives the minimum value, so $C = \frac{3}{4}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{3}{4}$.

P220. Let $a, b, c > 0$ such that $abc = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: $(a+b)(b+c)(c+a) \geq C(a+b+c-1)$

S220. $C = 4$

We will use the fact that $(a+b)(b+c)(c+a) \geq \frac{8}{9}(a+b+c)(a b+b c+c a)$. So, it is enough to prove that $\frac{2}{9}(a b+b c+c a)+\frac{1}{a+b+c} \geq 1$. Using the AM-GM Inequality, we can write\n\n$\n\\frac{2}{9}(a b+b c+c a)+\\frac{1}{a+b+c} \\geq 3 \\sqrt[3]{\\frac{(a b+b c+c a)^{2}}{81(a+b+c)}} \\geq 1\n$\n\nbecause\n\n$\n(a b+b c+c a)^{2} \\geq 3 a b c(a+b+c)=3(a+b+c)\n$\n\nEquality holds when $a = b = c = 1$, since then $abc = 1$ and $(a+b)(b+c)(c+a) = 8$, $a+b+c-1 = 2$, so $C = 4$ is achieved. This gives the maximum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 4$.”, “Using the identity $(a+b)(b+c)(c+a)=(a+b+c)(a b+b c+c a)-1$ we reduce the problem to the following one\n\n$\na b+b c+c a+\\frac{3}{a+b+c} \\geq 4\n$\n\nNow, we can apply the AM-GM Inequality in the following form\n\n$\na b+b c+c a+\\frac{3}{a+b+c} \\geq 4 \\sqrt[4]{\\frac{(a b+b c+c a)^{3}}{9(a+b+c)}}\n$\n\nAnd so it is enough to prove that\n\n$\n(a b+b c+c a)^{3} \\geq 9(a+b+c)\n$\n\nBut this is easy, because we clearly have $a b+b c+c a \geq 3$ and $(a b+b c+c a)^{2} \geq$ $3 a b c(a+b+c)=3(a+b+c)$.\n\nEquality holds when $a = b = c = 1$, since then $abc = 1$, $a+b+c = 3$, $ab+bc+ca = 3$, and $(a+b)(b+c)(c+a) = 8$, so $8 = 4(3-1)$. This gives the minimum value of $(a+b)(b+c)(c+a)$ for the given constraint, so $C = 4$ is the maximal constant.\n\nTherefore, the answer is $C = 4$.

P221. Let $a, b, c$ be positive real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: $a b \frac{a+c}{b+c}+b c \frac{b+a}{c+a}+c a \frac{c+b}{a+b} \geq C \sqrt{a b c(a+b+c)}$

S221. $C = \sqrt{3}$

Let $x=\frac{1}{b c}, y=\frac{1}{a c}, z=\frac{1}{a b}$ and $A=a c, B=a b, C=b c$.\nWe have\n$\n\\begin{aligned}\n\\frac{x}{y+z}(B+C) & =a b \\frac{a+c}{b+c}, \\quad \\frac{y}{z+x}(C+A)=b c \\frac{b+a}{c+a} \\quad \\text { and } \\\\\n\\frac{z}{x+y}(A+B) & =c a \\frac{c+b}{a+b}\n\\end{aligned}\n$\n\nUsing Corollary Let $a, b, c$ and $x, y, z$ be positive real numbers. Then\n$\n\\frac{x}{y+z}(b+c)+\\frac{y}{z+x}(c+a)+\\frac{z}{x+y}(a+b) \\geq \\sqrt{3(a b+b c+c a)}\n$\nand the previous identities we obtain\n$\n\\begin{aligned}\n& a b \\frac{a+c}{b+c}+b c \\frac{b+a}{c+a}+c a \\frac{c+b}{a+b} \\\\\n& \\quad=\\frac{x}{y+z}(B+C)+\\frac{y}{z+x}(C+A)+\\frac{z}{x+y}(A+B) \\\\\n& \\quad \\geq \\sqrt{3(A B+B C+C A)}=\\sqrt{3 a b c(a+b+c)}\n\\end{aligned}\n$\n\nEquality holds when $a = b = c$, since in this case all terms are equal and the inequality becomes an equality. This gives the minimum value of the left-hand side, so $C = \sqrt{3}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt{3}$.

P222. Let $a, b, x, y \in \mathbb{R}$ such that $a y - b x = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, x, y$ satisfying the given constraint: $a^{2}+b^{2}+x^{2}+y^{2}+a x+b y \geq C.$

S222. $C = \sqrt{3}$

Let us denote $u=a^{2}+b^{2}, v=x^{2}+y^{2}$ and $w=a x+b y$.\nThen\n$\n\\begin{aligned}\nu v & =\\left(a^{2}+b^{2}\\right)\\left(x^{2}+y^{2}\\right)=a^{2} x^{2}+a^{2} y^{2}+b^{2} x^{2}+b^{2} y^{2} \\\\\n& =a^{2} x^{2}+b^{2} y^{2}+2 a x b y+a^{2} y^{2}+b^{2} x^{2}-2 a x b y \\\\\n& =(a x+b y)^{2}+(a y-b x)^{2}=w^{2}+1\n\\end{aligned}\n$\n\nFrom the obvious inequality $(t \sqrt{3}+1)^{2} \geq 0$ we deduce\n$\n3 t^{2}+1 \\geq-2 t \\sqrt{3}\n$\ni.e.\n$\n4 t^{2}+4 \\geq 3-2 t \\sqrt{3}+t^{2}\n$\ni.e.\n$\n\\begin{equation*}\n4 t^{2}+4 \\geq(\\sqrt{3}-t)^{2} \\tag{1}\n\\end{equation*}\n$\n\nNow we have\n$\n(u+v)^{2} \\geq 4 u v=4\\left(w^{2}+1\\right) \\stackrel{(1)}{\\geq}(\\sqrt{3}-w)^{2}\n$\nfrom which we get $u+v \geq \sqrt{3}-w$, which is equivalent to $u+v+w \geq \sqrt{3}$.\n\nEquality holds when $a, b, x, y$ are such that $a y - b x = 1$ and $a x + b y = -\frac{\sqrt{3}}{2}$, with $a^2 + b^2 = x^2 + y^2 = \frac{\sqrt{3}}{2}$. This gives the minimum value of $a^2 + b^2 + x^2 + y^2 + a x + b y$, so $C = \sqrt{3}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt{3}$.

P223. Let $x, y, z$ be positive real numbers which satisfy the condition $x y + x z + y z + 2 x y z = 1.$ Find the largest constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint: $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - C(x + y + z) \geq \frac{(2z-1)^2}{z(2z+1)}.$ , where $z=\max {x, y, z}$.

S223. $C = 4$

Of course, if $z$ is the greatest from the numbers $x, y, z$, then $z \geq \frac{1}{2}$; we saw that\n\n$\n\\begin{aligned}\n\\frac{1}{x}+\\frac{1}{y}-4(x+y) & =(x+y)\\left(\\frac{1}{x y}-4\\right) \\geq \\frac{2}{2 z+1}(2 z+1) \\\\\n& =\\frac{2(2 z-1)(2 z+3)}{2 z+1}=4 z-\\frac{1}{z}+\\frac{(2 z-1)^{2}}{z(2 z+1)}\n\\end{aligned}\n$\n\nfrom where we get the inequality\n\n$\n\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}-4(x+y+z) \\geq \\frac{(2 z-1)^{2}}{z(2 z+1)}\n$\n\nOf course, in the right-hand side $z$ could be replaced by any of the three numbers which is $\geq \frac{1}{2}$ (two such numbers could be, surely there is one).\n\nEquality holds when $x = y = z = \frac{1}{2}$, since then the constraint is satisfied and the inequality becomes an equality. This gives the maximum value of $C$ for which the inequality always holds, namely $C = 4$.\n\nTherefore, the answer is $C = 4$.

P224. Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: $\sqrt{\frac{1}{a}-1} \sqrt{\frac{1}{b}-1} + \sqrt{\frac{1}{b}-1} \sqrt{\frac{1}{c}-1} + \sqrt{\frac{1}{c}-1} \sqrt{\frac{1}{a}-1} \geq C$

S224. $C = 6$

Let $a=x y, b=y z, c=z x$. Then $x y+y z+z x=1$ and we may take\n$\nx=\\tan \\frac{\\alpha}{2}, \\quad y=\\tan \\frac{\\beta}{2}, \\quad z=\\tan \\frac{\\gamma}{2}\n$\nwhere $\alpha, \beta, \gamma \in(0, \pi)$ and $\alpha+\beta+\gamma=\pi$.\nWe have\n$\n\\begin{aligned}\n\\sqrt{\\frac{1}{a}-1} \\sqrt{\\frac{1}{b}-1} & =\\sqrt{\\frac{(1-a)(1-b)}{a b}}=\\sqrt{\\frac{(1-x y)(1-y z)}{x y^{2} z}} \\\\\n& =\\sqrt{\\frac{(y z+z x)(z x+x y)}{x y^{2} z}}=\\sqrt{\\frac{(y+x)(z+y)}{y^{2}}}=\\frac{\\sqrt{1+y^{2}}}{y} \\\\\n& =\\frac{\\sqrt{1+\\tan ^{2} \\frac{\\beta}{2}}}{\\tan \\frac{\\beta}{2}}=\\frac{1}{\\sin \\frac{\\beta}{2}}\n\\end{aligned}\n$\n\nSimilarly we obtain\n$\n\\sqrt{\\frac{1}{b}-1} \\sqrt{\\frac{1}{c}-1}=\\frac{1}{\\sin \\frac{\\gamma}{2}} \\quad \\text { and } \\quad \\sqrt{\\frac{1}{c}-1} \\sqrt{\\frac{1}{a}-1}=\\frac{1}{\\sin \\frac{\\alpha}{2}}\n$\n\nNow the given inequality becomes\n$\n\\frac{1}{\\sin \\frac{\\alpha}{2}}+\\frac{1}{\\sin \\frac{\\beta}{2}}+\\frac{1}{\\sin \\frac{\\gamma}{2}} \\geq 6\n$\n\nBy $A M \geq H M$ we have\n$\n\\frac{1}{\\sin \\frac{\\alpha}{2}}+\\frac{1}{\\sin \\frac{\\beta}{2}}+\\frac{1}{\\sin \\frac{\\gamma}{2}} \\geq \\frac{9}{\\sin \\frac{\\alpha}{2}+\\sin \\frac{\\beta}{2}+\\sin \\frac{\\gamma}{2}}\n$\n\nSo we need to prove that $\sin \frac{\alpha}{2}+\sin \frac{\beta}{2}+\sin \frac{\gamma}{2} \leq \frac{3}{2}$.\n\nEquality occurs if and only if $\alpha=\beta=\gamma=\pi / 3$, i.e. $a=b=c=\frac{1}{3}$. In this case, the minimum value of the expression is achieved, so $C = 6$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 6$.

P225. Let $x, y, z, t \in \mathbb{R}^{+}$. Find the smallest constant $C$ such that the following inequality holds for all positive $x, y, z, t$: $x^{4}+y^{4}+z^{4}+t^{4}+C x y z t \geq x^{2} y^{2}+y^{2} z^{2}+z^{2} t^{2}+t^{2} x^{2}+x^{2} z^{2}+y^{2} t^{2}$

S225. $C = 2$

Clearly, it is enough to prove the inequality if $x y z t=1$ and so the problem becomes\n\nIf $a, b, c, d$ have product 1 , then $a^{2}+b^{2}+c^{2}+d^{2}+2 \geq a b+b c+c d+d a+a c+b d$. Let $d$ the minimum among $a, b, c, d$ and let $m=\sqrt[3]{a b c}$. We will prove that\n$a^{2}+b^{2}+c^{2}+d^{2}+2-(a b+b c+c d+d a+a c+b d) \geq d^{2}+3 m^{2}+2-\left(3 m^{2}+3 m d\right)$, which is in fact\n\n$\na^{2}+b^{2}+c^{2}-a b-b c-c a \\geq d(a+b+c-3 \\sqrt[3]{a b c})\n$\n\nBecause $d \leq \sqrt[3]{a b c}$, proving this first inequality comes down to the inequality\n\n$\na^{2}+b^{2}+c^{2}-a b-b c-c a \\geq \\sqrt[3]{a b c}(a+b+c-3 \\sqrt[3]{a b c})\n$\n\nTake $u=\frac{a}{\sqrt[3]{a b c}}, v=\frac{b}{\sqrt[3]{a b c}}, w=\frac{c}{\sqrt[3]{a b c}}$. Using problem 74 , we find that\n\n$\nu^{2}+v^{2}+w^{2}+3 \\geq u+v+w+u v+v w+w u\n$\n\nwhich is exactly $a^{2}+b^{2}+c^{2}-a b-b c-c a \geq \sqrt[3]{a b c}(a+b+c-3 \sqrt[3]{a b c})$. Thus, it remains to prove that $d^{2}+2 \geq 3 m d \Leftrightarrow d^{2}+2 \geq 3 \sqrt[3]{d^{2}}$, which is clear.\n\nEquality holds when $x = y = z = t = 1$, since then both sides of the original inequality are equal: $4 + 2 \cdot 1 = 6$ and $6$. This gives the minimum value of $C$, so $C = 2$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 2$.

P226. For any positive real numbers $x, y$ and any positive integers $m, n$, there exists a constant $C$ such that the following inequality holds: $C\left(x^{m+n} + y^{m+n}\right) + (m+n-1)\left(x^m y^n + x^n y^m\right) \geq m n\left(x^{m+n-1} y + y^{m+n-1} x\right).$ Determine the optimal value of $C$.

S226. $C = (n-1)(m-1)$

We transform the inequality as follows:\n\n$\n\\begin{gathered}\nm n(x-y)\\left(x^{m+n-1}-y^{m+n-1}\\right) \\geq(m+n-1)\\left(x^{m}-y^{m}\\right)\\left(x^{n}-y^{n}\\right) \\Leftrightarrow \\\\\n\\Leftrightarrow \\frac{x^{m+n-1}-y^{m+n-1}}{(m+n-1)(x-y)} \\geq \\frac{x^{m}-y^{m}}{m(x-y)} \\cdot \\frac{x^{n}-y^{n}}{n(x-y)}\n\\end{gathered}\n$\n\n(we have assumed that $x>y$ ). The last relation can also be written\n\n$\n(x-y) \\int_{y}^{x} t^{m+n-2} d t \\geq \\int_{y}^{x} t^{m-1} d t \\cdot \\int_{y}^{x} t^{n-1} d t\n$\n\nand this follows from Chebyshev’s Inequality for integrals.\n\nEquality holds when $x = y$, since all terms become equal and the inequality becomes an equality. This gives the minimum value of $C$, so the optimal value is $C = (n-1)(m-1)$.\n\nTherefore, the answer is $C = (n-1)(m-1)$.

P227. Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: $\left(a^{a}+b^{a}+c^{a}\right)\left(a^{b}+b^{b}+c^{b}\right)\left(a^{c}+b^{c}+c^{c}\right) \geq C(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c})^{3}$

S227. $C = 1$

By Hölder’s inequality we obtain\n$\n\\left(a^{a}+b^{a}+c^{a}\\right)^{\\frac{1}{3}}\\left(a^{b}+b^{b}+c^{b}\\right)^{\\frac{1}{3}}\\left(a^{c}+b^{c}+c^{c}\\right)^{\\frac{1}{3}} \\geq a^{\\frac{a+b+c}{3}}+b^{\\frac{a+b+c}{3}}+c^{\\frac{a+b+c}{3}}\n$\n\nSince $a+b+c=1$, the conclusion follows.\n\nEquality holds when $a = b = c = \frac{1}{3}$, in which case both sides of the inequality are equal. This gives the minimum value of the left side, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P228. Let $a, b, c \in \mathbb{R}^{+}$ such that $abc = 1$. Find the smallest constant $C$ such that the following inequality holds for all $a, b, c$: $\frac{1}{1+a+b}+\frac{1}{1+b+c}+\frac{1}{1+c+a} \leq C \left( \frac{1}{2+a}+\frac{1}{2+b}+\frac{1}{2+c} \right)$

S228. $C = 1$

Let $x=a+b+c$ and $y=a b+b c+c a$. Using brute-force, it is easy to see that the left hand side is $\frac{x^{2}+4 x+y+3}{x^{2}+2 x+y+x y}$, while the right hand side is $\frac{12+4 x+y}{9+4 x+2 y}$. Now, the inequality becomes\n\n$\n\\frac{x^{2}+4 x+y+3}{x^{2}+2 x+y+x y}-1 \\leq \\frac{12+4 x+y}{9+4 x+2 y}-1 \\Leftrightarrow \\frac{2 x+3-x y}{x^{2}+2 x+y+x y} \\leq \\frac{3-y}{9+4 x+2 y}\n$\n\nFor the last inequality, we clear denominators. Then using the inequalities $x \geq 3, y \geq 3, x^{2} \geq 3 y$, we have\n\n$\n\\frac{5}{3} x^{2} y \\geq 5 x^{2}, \\frac{x^{2} y}{3} \\geq y^{2}, x y^{2} \\geq 9 x, 5 x y \\geq 15 x, x y \\geq 3 y \\text { and } x^{2} y \\geq 27\n$\n\nSumming up these inequalities, the desired inequality follows.\n\nEquality holds when $a = b = c = 1$, since then $abc = 1$ and both sides of the original inequality are equal. This gives the minimum value of $C$, so $C = 1$ is the smallest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P229. Let $a, b, c$ be positive real numbers such that $a + b + c = 3$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: $\left(1 + a + a^2\right)\left(1 + b + b^2\right)\left(1 + c + c^2\right) \geq C(ab + bc + ca).$

S229. $C = 9$

Let us denote $x=a+b+c=3, y=a b+b c+c a, z=a b c$.\nNow the given inequality can be rewritten as\n$\nz^{2}-2 z-2 x z+z(x+y)+x^{2}+x+y^{2}-y+3 x y+1 \\geq 9 y\n$\ni.e.\n$\n(z-1)^{2}-(z-1)(x-y)+(x-y)^{2} \\geq 0\n$\nwhich is obviously true. Equality holds iff $a=b=c=1$.\n\nEquality is achieved when $a = b = c = 1$, in which case $ab + bc + ca = 3$ and $\left(1 + a + a^2\right)^3 = 27$, so the inequality becomes $27 \geq 9 \times 3 = 27$. Thus, $C = 9$ is the largest constant for which the inequality always holds, and this gives the minimum value of $C$.\n\nTherefore, the answer is $C = 9$.

P230. Let $a, b, c \in (-1, 1)$ be real numbers such that $a b + b c + a c = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: $C \sqrt[3]{\left(1-a^{2}\right)\left(1-b^{2}\right)\left(1-c^{2}\right)} \leq 1 + (a + b + c)^2$

S230. $C = 6$

Since $a, b, c \in(-1,1)$ we have $1-a^{2}, 1-b^{2}, 1-c^{2}>0$.\nBy $A M \geq G M$ we get\n$\n\\begin{aligned}\n6 \\sqrt[3]{\\left(1-a^{2}\\right)\\left(1-b^{2}\\right)\\left(1-c^{2}\\right)} & =2 \\cdot 3 \\sqrt[3]{\\left(1-a^{2}\\right)\\left(1-b^{2}\\right)\\left(1-c^{2}\\right)} \\\\\n& \\leq 2\\left(1-a^{2}+1-b^{2}+1-c^{2}\\right) \\\\\n& =2\\left(3-\\left(a^{2}+b^{2}+c^{2}\\right)\\right) \\\\\n& =6-2\\left(a^{2}+b^{2}+c^{2}\\right)\n\\end{aligned}\n$\n\nWe’ll show that\n$\n6-2\\left(a^{2}+b^{2}+c^{2}\\right) \\leq 1+(a+b+c)^{2}\n$\n\nThis inequality is equivalent to\n$\n6-2\\left(a^{2}+b^{2}+c^{2}\\right) \\leq 1+a^{2}+b^{2}+c^{2}+2\n$\ni.e.\n$\n3 \\leq 3\\left(a^{2}+b^{2}+c^{2}\\right)\n$\ni.e.\n$\na^{2}+b^{2}+c^{2} \\geq 1\n$\nwhich is true since $a^{2}+b^{2}+c^{2} \geq a b+b c+a c=1$.\n\nEquality holds when $a = b = c = \pm \frac{1}{\sqrt{3}}$, since then $ab + bc + ca = 1$ and $a^2 + b^2 + c^2 = 1$. In this case, the inequality becomes equality, so $C = 6$ is the largest possible value.\n\nTherefore, the answer is $C = 6$.

P231. Let $x, y, z > 0$ satisfy the condition $x + y + z = xyz$. Find the maximal constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given condition: $xy + xz + yz \geq C + \sqrt{x^2 + 1} + \sqrt{y^2 + 1} + \sqrt{z^2 + 1}$

S231. $C = 3$

Another improvement is as follows. Start from\n\n$\n\\frac{1}{x^{2}}+\\frac{1}{y^{2}}+\\frac{1}{z^{2}} \\geq \\frac{1}{x y}+\\frac{1}{x z}+\\frac{1}{y z}=1 \\Rightarrow x^{2} y^{2}+x^{2} z^{2}+y^{2} z^{2} \\geq x^{2} y^{2} z^{2}\n$\n\nwhich is equivalent to\n\n$\n(x y+x z+y z)^{2} \\geq 2 x y z(x+y+z)+x^{2} y^{2} z^{2}=3(x+y+z)^{2}\n$\n\nFurther on,\n\n$\n\\begin{aligned}\n& (x y+x z+y z-3)^{2}=(x y+x z+y z)^{2}-6(x y+x z+y z)+9 \\geq \\\\\n& \\geq 3(x+y+z)^{2}-6(x y+x z+y z)+9=3\\left(x^{2}+y^{2}+z^{2}\\right)+9\n\\end{aligned}\n$\n\nso that\n\n$\nx y+x z+y z \\geq 3+\\sqrt{3\\left(x^{2}+y^{2}+z^{2}\\right)+9}\n$\n\nBut\n\n$\n\\sqrt{3\\left(x^{2}+y^{2}+z^{2}\\right)+9} \\geq \\sqrt{x^{2}+1}+\\sqrt{y^{2}+1}+\\sqrt{z^{2}+1}\n$\n\nis a consequence of the Cauchy-Schwarz Inequality and we have a second improvement and proof for the desired inequality:\n\n$\n\\begin{aligned}\nx y+x z+y z & \\geq 3+\\sqrt{3\\left(x^{2}+y^{2}+z^{2}\\right)+9} \\geq \\\\\n& \\geq 3+\\sqrt{x^{2}+1}+\\sqrt{y^{2}+1}+\\sqrt{z^{2}+1}\n\\end{aligned}\n$\n\nEquality holds when $x = y = z = \sqrt{3}$, since then $x + y + z = 3\sqrt{3}$ and $xyz = (\sqrt{3})^3 = 3\sqrt{3}$, so the condition is satisfied. In this case, $xy + xz + yz = 3 \cdot 3 = 9$ and $\sqrt{x^2+1} = \sqrt{3+1} = 2$ for each variable, so $C = 9 - 3 \times 2 = 3$. This gives the minimum value of $xy + xz + yz - (\sqrt{x^2+1} + \sqrt{y^2+1} + \sqrt{z^2+1})$, so the maximal constant $C$ is $3$.\n\nTherefore, the answer is $C = 3$. “, “We have\n\n$\nx y z=x+y+z \\geq 2 \\sqrt{x y}+z \\Rightarrow z(\\sqrt{x y})^{2}-2 \\sqrt{x y}-z \\geq 0\n$\n\nBecause the positive root of the trinomial $z t^{2}-2 t-z$ is\n\n$\n\\frac{1+\\sqrt{1+z^{2}}}{z}\n$\n\nwe get from here\n\n$\n\\sqrt{x y} \\geq \\frac{1+\\sqrt{1+z^{2}}}{z} \\Leftrightarrow z \\sqrt{x y} \\geq 1+\\sqrt{1+z^{2}}\n$\n\nOf course, we have two other similar inequalities. Then,\n\n$\n\\begin{aligned}\nx y+x z+y z & \\geq x \\sqrt{y z}+y \\sqrt{x z}+z \\sqrt{x y} \\geq \\\\\n& \\geq 3+\\sqrt{x^{2}+1}+\\sqrt{y^{2}+1}+\\sqrt{z^{2}+1}\n\\end{aligned}\n$\n\nand we have both a proof of the given inequality, and a little improvement of it.\n\nEquality holds when $x = y = z = 1$, since then $x + y + z = 3 = xyz$, and $xy + yz + zx = 3$, $\sqrt{x^2+1} + \sqrt{y^2+1} + \sqrt{z^2+1} = 3\sqrt{2}$, so the inequality becomes $3 \geq 3 + 3\sqrt{2}$, which is tight for $C = 3$. This gives the minimum value of $C$ for which the inequality always holds.\n\nTherefore, the answer is $C = 3$.

P232. Let $a, b, c \in \mathbb{R}^{+}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: $\frac{a}{b+2 c}+\frac{b}{c+2 a}+\frac{c}{a+2 b} \geq C$

S232. $C = 1$

Applying the Cauchy-Schwarz inequality we get\n$\n\\begin{aligned}\n& \\left(\\frac{a}{b+2 c}+\\frac{b}{c+2 a}+\\frac{c}{a+2 b}\\right)(a(b+2 c)+b(c+2 a)+c(a+2 b)) \\\\\n& \\quad \\geq(a+b+c)^{2}\n\\end{aligned}\n$\nhence\n$\n\\frac{a}{b+2 c}+\\frac{b}{c+2 a}+\\frac{c}{a+2 b} \\geq \\frac{(a+b+c)^{2}}{3(a b+b c+c a)}\n$\n\nSo it suffices to show that\n$\n\\frac{(a+b+c)^{2}}{3(a b+b c+c a)} \\geq 1, \\quad \\text { i.e. } \\quad(a+b+c)^{2} \\geq 3(a b+b c+c a)\n$\nwhich is equivalent to $a^{2}+b^{2}+c^{2} \geq a b+b c+c a$, and clearly holds.\nEquality occurs iff $a=b=c$.\n\nEquality holds when $a = b = c$, in which case $\frac{a}{b+2c} + \frac{b}{c+2a} + \frac{c}{a+2b} = 1$. This gives the minimum value of the expression, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P233. Let $a, b, c$ be the lengths of the sides of a triangle, such that $a + b + c = 3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$: $a^{2} + b^{2} + c^{2} + \frac{4 a b c}{3} \geq C.$

S233. $C = \frac{13}{3}$

Let $a=x+y, b=y+z$ and $c=z+x$.\nSo we have $x+y+z=\frac{3}{2}$ and since $A M \geq G M$ we get $x y z \leq\left(\frac{x+y+z}{3}\right)^{3}=\frac{1}{8}$.\n\nNow we obtain\n$\n\\begin{aligned}\na^{2} & +b^{2}+c^{2}+\\frac{4 a b c}{3} \\\\\n& =\\frac{\\left(a^{2}+b^{2}+c^{2}\\right)(a+b+c)+4 a b c}{3} \\\\\n& =\\frac{2\\left((x+y)^{2}+(y+z)^{2}+(z+x)^{2}\\right)(x+y+z)+4(x+y)(y+z)(z+x)}{3} \\\\\n& =\\frac{4}{3}\\left((x+y+z)^{3}-x y z\\right) \\geq \\frac{4}{3}\\left(\\left(\\frac{3}{2}\\right)^{3}-\\frac{1}{8}\\right)=\\frac{13}{3}\n\\end{aligned}\n$\n\nEquality holds when $x = y = z$, which means $a = b = c = 1$. In this case, the minimum value of the expression is achieved, so $C = \frac{13}{3}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{13}{3}$.

P234. Let $a, b, c \in \mathbb{R}^{+}$ with $a^{2}+b^{2}+c^{2}=3$. Determine the largest constant $C$ such that the following inequality holds for all $a, b, c$: $\frac{a b}{c}+\frac{b c}{a}+\frac{c a}{b} \geq C.$

S234. $C = 3$

The given inequality is equivalent to\n$\n\\begin{aligned}\n& \\left(\\frac{a b}{c}+\\frac{b c}{a}+\\frac{c a}{b}\\right)^{2} \\geq 9 \\\\\n& \\quad \\Leftrightarrow \\quad \\frac{a^{2} b^{2}}{c^{2}}+\\frac{b^{2} c^{2}}{a^{2}}+\\frac{c^{2} a^{2}}{b^{2}}+2\\left(a^{2}+b^{2}+c^{2}\\right) \\geq 3\\left(a^{2}+b^{2}+c^{2}\\right)\n\\end{aligned}\n$\ni.e.\n$\n\\frac{a^{2} b^{2}}{c^{2}}+\\frac{b^{2} c^{2}}{a^{2}}+\\frac{c^{2} a^{2}}{b^{2}} \\geq a^{2}+b^{2}+c^{2}\n$\n\nFurthermore, applying $A M \geq G M$ we get\n$\n\\frac{a^{2} b^{2}}{c^{2}}+\\frac{b^{2} c^{2}}{a^{2}} \\geq 2 b^{2}, \\quad \\frac{b^{2} c^{2}}{a^{2}}+\\frac{c^{2} a^{2}}{b^{2}} \\geq 2 c^{2}, \\quad \\frac{a^{2} b^{2}}{c^{2}}+\\frac{c^{2} a^{2}}{b^{2}} \\geq 2 a^{2}\n$\n\nAfter adding these inequalities we obtain\n$\n\\frac{a^{2} b^{2}}{c^{2}}+\\frac{b^{2} c^{2}}{a^{2}}+\\frac{c^{2} a^{2}}{b^{2}} \\geq a^{2}+b^{2}+c^{2}\n$\nand we are done.\n\nEquality holds when $a = b = c = 1$, since then $a^2 + b^2 + c^2 = 3$ and $\frac{ab}{c} + \frac{bc}{a} + \frac{ca}{b} = 3$. This gives the minimum value of the expression, so the largest constant $C$ for which the inequality always holds is $C = 3$.\n\nTherefore, the answer is $C = 3$.

P235. Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $a, b, c$: $\frac{a}{a + bc} + \frac{b}{b + ca} + \frac{\sqrt{abc}}{c + ab} \leq C.$

S235. $C = 1 + \frac{3\sqrt{3}}{4}$

Since $a+b+c=1$ we use the following substitutions $a=xy, b=yz, c=zx$, where $x, y, z>0$ and the given inequality becomes\n$\n\\frac{xy}{xy+(yz)(zx)}+\\frac{yz}{yz+(zx)(xy)}+\\frac{xyz}{zx+(xy)(yz)} \\leq 1+\\frac{3 \\sqrt{3}}{4}\n$\ni.e.\n$\n\\begin{equation*}\n\\frac{1}{1+z^{2}}+\\frac{1}{1+x^{2}}+\\frac{y}{1+y^{2}} \\leq 1+\\frac{3 \\sqrt{3}}{4} \\tag{1}\n\\end{equation*}\n$\nwhere $xy+yz+zx=1$.\n\nSince $xy+yz+zx=1$ according to Case 3 we may set $x=\tan \frac{\alpha}{2}, y=\tan \frac{\beta}{2}, z=\tan \frac{\gamma}{2}$ where $\alpha, \beta, \gamma \in(0, \pi)$, and $\alpha+\beta+\gamma=\pi$.\n\nThen inequality (1) becomes\n$\n\\frac{1}{1+\\tan ^{2} \\frac{\\gamma}{2}}+\\frac{1}{1+\\tan ^{2} \\frac{\\alpha}{2}}+\\frac{\\tan \\frac{\\beta}{2}}{1+\\tan ^{2} \\frac{\\beta}{2}} \\leq 1+\\frac{3 \\sqrt{3}}{4}\n$\ni.e.\n$\n\\cos ^{2} \\frac{\\gamma}{2}+\\cos ^{2} \\frac{\\alpha}{2}+\\frac{\\sin \\beta}{2} \\leq 1+\\frac{3 \\sqrt{3}}{4}\n$\n\nUsing the trigonometric identity $\cos x=2 \cos ^{2} \frac{x}{2}-1$ the last inequality becomes\n$\n\\frac{\\cos \\gamma+1}{2}+\\frac{\\cos \\alpha+1}{2}+\\frac{\\sin \\beta}{2} \\leq 1+\\frac{3 \\sqrt{3}}{4}\n$\ni.e.\n$\n\\begin{equation*}\n\\cos \\gamma+\\cos \\alpha+\\sin \\beta \\leq \\frac{3 \\sqrt{3}}{2} \\tag{2}\n\\end{equation*}\n$\n\nWe have\n$\n\\begin{aligned}\n\\cos \\alpha+\\cos \\gamma+\\sin \\beta= & \\cos \\alpha+\\cos \\gamma+\\sin (\\pi-(\\alpha+\\gamma)) \\\\\n= & \\frac{2}{\\sqrt{3}}\\left(\\frac{\\sqrt{3}}{2} \\cos \\alpha+\\frac{\\sqrt{3}}{2} \\cos \\gamma\\right) \\\\\n& +\\frac{1}{\\sqrt{3}}(\\sqrt{3} \\sin \\alpha \\cos \\gamma+\\sqrt{3} \\cos \\alpha \\sin \\gamma) \\\\\n\\leq & \\frac{1}{\\sqrt{3}}\\left(\\frac{3}{4}+\\cos ^{2} \\alpha+\\frac{3}{4}+\\cos ^{2} \\gamma\\right) \\\\\n= & \\frac{1}{2 \\sqrt{3}}\\left(3 \\sin ^{2} \\alpha+\\cos ^{2} \\gamma+\\cos ^{2} \\alpha+3 \\sin ^{2} \\gamma\\right) \\\\\n= & \\frac{\\sqrt{3}}{2}+\\frac{\\sqrt{3}}{2}\\left(\\cos ^{2} \\alpha+\\sin ^{2} \\alpha\\right)+\\frac{\\sqrt{3}}{2}\\left(\\cos ^{2} \\gamma+\\sin ^{2} \\gamma\\right) \\\\\n= & \\frac{3 \\sqrt{3}}{2} .\n\\end{aligned}\n$\n\nEquality in the above inequalities is achieved when $\alpha = \gamma = \frac{\pi}{3}$ and $\beta = \frac{\pi}{3}$, which corresponds to $a = b = c = \frac{1}{3}$. In this case, the left-hand side attains its maximum value, so $C = 1 + \frac{3\sqrt{3}}{4}$ is the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1 + \frac{3\sqrt{3}}{4}$.

P236. Let $x, y, z$ be positive real numbers. Find the minimal constant $C$ such that the following inequality holds for all $x, y, z \in \mathbb{R}^{+}$: $\frac{x}{x+\sqrt{(x+y)(x+z)}}+\frac{y}{y+\sqrt{(y+z)(y+x)}}+\frac{z}{z+\sqrt{(z+x)(z+y)}} \leq C$

S236. $C = 1$

From Huygens Inequality we have $\sqrt{(x+y)(x+z)} \geq x+\sqrt{y z}$ and using this inequality for the similar ones we get\n\n$\n\\begin{aligned}\n\\frac{x}{x+\\sqrt{(x+y)(x+z)}} & +\\frac{y}{y+\\sqrt{(y+z)(y+x)}}+\\frac{z}{z+\\sqrt{(z+x)(z+y)}} \\leq \\\\\n& \\leq \\frac{x}{2 x+\\sqrt{y z}}+\\frac{y}{2 y+\\sqrt{z x}}+\\frac{z}{2 z+\\sqrt{x y}}\n\\end{aligned}\n$\n\nNow, we denote with $a=\frac{\sqrt{y z}}{x}, b=\frac{\sqrt{z x}}{y}, c=\frac{\sqrt{x y}}{z}$ and the inequality becomes\n\n$\n\\frac{1}{2+a}+\\frac{1}{2+b}+\\frac{1}{2+c} \\leq 1\n$\n\nFrom the above notations we can see that $a b c=1$, so the last inequality becomes after clearing the denominators $a b+b c+c a \geq 3$, which follows from the AM-GM Inequality.\n\nEquality holds when $x = y = z$, in which case $a = b = c = 1$ and $\frac{1}{2+1} + \frac{1}{2+1} + \frac{1}{2+1} = 1$. This gives the maximum value of the original expression, so $C = 1$ is the minimal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$. “, “We have $(x+y)(x+z)=x y+\left(x^{2}+y z\right)+x z \geq x y+2 x \sqrt{y z}+x z=(\sqrt{x y}+\sqrt{x z})^{2}$. Hence\n\n$\n\\sum \\frac{x}{x+\\sqrt{(x+y)(x+z)}} \\leq \\sum \\frac{x}{x+\\sqrt{x y}+\\sqrt{x z}}\n$\n\nBut\n\n$\n\\sum \\frac{x}{x+\\sqrt{x y}+\\sqrt{x z}}=\\sum \\frac{\\sqrt{x}}{\\sqrt{x}+\\sqrt{y}+\\sqrt{z}}=1\n$\n\nand this solves the problem.\n\nEquality holds when $x = y = z$, since then each term becomes $\frac{x}{x + x + x} = \frac{1}{3}$, and the sum is $1$. This gives the maximum value of the sum, so $C = 1$ is minimal such that the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P237. Let $a, b, c$ be non-negative real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}^{+}$: $\frac{a}{4 b^{2}+b c+4 c^{2}}+\frac{b}{4 c^{2}+c a+4 a^{2}}+\frac{c}{4 a^{2}+a b+4 b^{2}} \geq C \cdot \frac{1}{a+b+c}$

S237. $C = 1$

By the Cauchy-Schwarz inequality we have\n$\n\\begin{gathered}\n\\frac{a}{4 b^{2}+b c+4 c^{2}}+\\frac{b}{4 c^{2}+c a+4 a^{2}}+\\frac{c}{4 a^{2}+a b+4 b^{2}} \\\\\n\\geq \\frac{(a+b+c)^{2}}{4 a\\left(b^{2}+c^{2}\\right)+4 b\\left(c^{2}+a^{2}\\right)+4 c\\left(a^{2}+b^{2}\\right)+3 a b c}\n\\end{gathered}\n$\n\nSo we need to prove that\n$\n\\frac{(a+b+c)^{2}}{4 a\\left(b^{2}+c^{2}\\right)+4 b\\left(c^{2}+a^{2}\\right)+4 c\\left(a^{2}+b^{2}\\right)+3 a b c} \\geq \\frac{1}{a+b+c}\n$\nwhich is equivalent to\n$\n(a+b+c)^{3} \\geq 4 a\\left(b^{2}+c^{2}\\right)+4 b\\left(c^{2}+a^{2}\\right)+4 c\\left(a^{2}+b^{2}\\right)+3 a b c\n$\ni.e.\n$\na^{3}+b^{3}+c^{3}+3 a b c \\geq a\\left(b^{2}+c^{2}\\right)+b\\left(c^{2}+a^{2}\\right)+c\\left(a^{2}+b^{2}\\right)\n$\nwhich is Schur’s inequality.\n\nEquality holds when $a = b = c$, for example at $a = b = c > 0$. In this case, both sides of the inequality are equal, so the minimum value of the left side is achieved and $C = 1$ is the largest possible constant.\n\nTherefore, the answer is $C = 1$.

P238. Let $a, b, c \in \mathbb{R}$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$: $a^{2}+b^{2}+c^{2} \geq C(a b+b c+c a)$

S238. $C = 1$

Since $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$ we deduce\n$\n2\\left(a^{2}+b^{2}+c^{2}\\right) \\geq 2(a b+b c+c a) \\Leftrightarrow a^{2}+b^{2}+c^{2} \\geq a b+b c+c a\n$\n\nEquality occurs if and only if $a = b = c$, in which case both sides are equal. This gives the minimum value of $a^2 + b^2 + c^2 - C(ab + bc + ca)$, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P239. Let $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1}+a_{2}+\cdots+a_{n}=1$. Find the largest constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$: $\frac{a_{1}}{\sqrt{1-a_{1}}}+\frac{a_{2}}{\sqrt{1-a_{2}}}+\cdots+\frac{a_{n}}{\sqrt{1-a_{n}}} \geq C$

S239. $C = \sqrt{\frac{n}{n-1}}$

Let us denote\n$\n\\begin{aligned}\n& A=\\frac{a_{1}}{\\sqrt{1-a_{1}}}+\\frac{a_{2}}{\\sqrt{1-a_{2}}}+\\cdots+\\frac{a_{n}}{\\sqrt{1-a_{n}}} \\\\\n& B=a_{1}\\left(1-a_{1}\\right)+a_{2}\\left(1-a_{2}\\right)+\\cdots+a_{n}\\left(1-a_{n}\\right)\n\\end{aligned}\n$\n\nBy Hölder’s inequality we have\n$\n\\begin{equation*}\nA^{2} B \\geq\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{3}=1 \\tag{1}\n\\end{equation*}\n$\n\nApplying $QM \geq AM$ we deduce\n$\n\\begin{equation*}\nB=1-\\left(a_{1}^{2}+a_{2}^{2}+\\cdots+a_{n}^{2}\\right) \\leq 1-\\frac{\\left(a_{1}+a_{2}+\\cdots+a_{n}\\right)^{2}}{n}=\\frac{n-1}{n} \\tag{2}\n\\end{equation*}\n$\n\nBy (1) and (2) we obtain\n$\n\\frac{n-1}{n} \\cdot A^{2} \\geq A^{2} B \\geq 1, \\quad \\text { i.e. } \\quad A \\geq \\sqrt{\\frac{n}{n-1}}\n$\n\nEquality holds when $a_{1} = a_{2} = \cdots = a_{n} = \frac{1}{n}$, since then both Hölder’s and QM-AM inequalities become equalities. This gives the minimum value of $A$, so $C = \sqrt{\frac{n}{n-1}}$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = \sqrt{\frac{n}{n-1}}$.

P240. Let $a, b, c$ be real numbers. Find the largest constant $C$ such that the following inequality holds for all $a, b, c \in \mathbb{R}$: $\left(a^{2}+b^{2}+c^{2}\right)^{2} \geq C\left(a^{3} b+b^{3} c+c^{3} a\right)$

S240. $C = 3$

By the well-known inequality $(x+y+z)^{2} \geq 3(xy+yz+zx)$ for\n$\nx = a^{2} + b c - a b, \\quad y = b^{2} + c a - b c, \\quad z = c^{2} + a b - c a\n$\nwe obtain the required inequality.\n\nEquality holds when $a = b = c$, in which case both sides of the inequality are equal. This gives the maximum value of $C$ for which the inequality always holds, namely $C = 3$.\n\nTherefore, the answer is $C = 3$.

P241. Let $a, b, c$ be positive real numbers such that $a+b+c=1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: $(a+b)^{2}(1+2 c)(2 a+3 c)(2 b+3 c) \geq C a b c$

S241. $C = 54$

The given inequality can be rewritten as follows\n$\n(a+b)^{2}(1+2 c)\\left(2+3 \\frac{c}{a}\\right)\\left(2+3 \\frac{c}{b}\\right) \\geq 54 c\n$\n\nBy the Cauchy-Schwarz inequality and $A M \geq G M$ we have\n$\n\\begin{aligned}\n\\left(2+3 \\frac{c}{a}\\right)\\left(2+3 \\frac{c}{b}\\right) & \\geq\\left(2+\\frac{3 c}{\\sqrt{a b}}\\right)^{2} \\geq\\left(2+\\frac{6 c}{a+b}\\right)^{2}=\\frac{(2(a+b)+6 c)^{2}}{(a+b)^{2}} \\\\\n& =\\frac{4(1+2 c)^{2}}{(a+b)^{2}}\n\\end{aligned}\n$\n\nThen we have\n$\n\\begin{aligned}\n(a+b)^{2}(1+2 c)\\left(2+3 \\frac{c}{a}\\right)\\left(2+3 \\frac{c}{b}\\right) & \\geq(a+b)^{2}(1+2 c) \\frac{4(1+2 c)^{2}}{(a+b)^{2}} \\\\\n& =4(1+2 c)^{3}\n\\end{aligned}\n$\nand it remains to prove that\n$\n4(1+2 c)^{3} \\geq 54 c, \\quad \\text { i.e. } \\quad(1+2 c)^{3} \\geq \\frac{27 c}{2}\n$\n\nBy the $A M \geq G M$ inequality we have\n$\n(1+2 c)^{3}=\\left(\\frac{1}{2}+\\frac{1}{2}+2 c\\right)^{3} \\geq 27 \\cdot \\frac{1}{2} \\cdot \\frac{1}{2} \\cdot 2 c=\\frac{27 c}{2}\n$\nas required.\nEquality occurs iff $a=b=\frac{3}{8},\ c=\frac{1}{4}$, that is, when $a$ and $b$ are equal and $c$ is $\frac{1}{4}$. This gives the maximum value of $C$ for which the inequality always holds, so $C=54$ is maximal.\n\nTherefore, the answer is $C = 54$.

P242. Let $a, b, c > 0$ be real numbers. Determine the maximal constant $C$ such that the following inequality holds for all $a, b, c$: $(a b + b c + c a) \left( \frac{1}{(a + b)^2} + \frac{1}{(b + c)^2} + \frac{1}{(c + a)^2} \right) \geq C.$

S242. $C = \frac{9}{4}$

We can rewrite the given inequality in the following form\n$\n\\begin{aligned}\n& f(a+b+c, a b+b c+c a, a b c) \\\\\n&= 9((a+b)(b+c)(c+a))^{2} \\\\\n&-4(a b+b c+c a)\\left((a+b)^{2}(b+c)^{2}+(b+c)^{2}(c+a)^{2}+(c+a)^{2}(a+b)^{2}\\right) \\\\\n&= k(a b c)^{2}+m a b c+n\n\\end{aligned}\n$\nwhere $k \geq 0$ and $k, m, n$ are quantities containing constants or $a+b+c, a b+$ $b c+c a, a b c$, which we also consider as constants, i.e. in the form as a sixth-degree symmetric polynomial with variables $a, b, c$ and a second-degree polynomial with variable $a b c$ and positive coefficients.\n\nLet us explain this:\nThe expression $(a+b)(b+c)(c+a)$ has the form $k a b c+m$ so it follows that $9((a+b)(b+c)(c+a))^{2}$ has the form $k^{2}(a b c)^{2}+m a b c+n$.\n\nFurthermore\n$4(a b+b c+c a)\left((a+b)^{2}(b+c)^{2}+(b+c)^{2}(c+a)^{2}+(c+a)^{2}(a+b)^{2}\right)=4 k A$,\nwhere $k=a b+b c+c a$, and $A$ is a fourth-degree polynomial and also has the form $k a b c+m$.\n\nTherefore the expression of the left side of $f(a+b+c, a b+b c+c a, a b c)$ has the form $k(a b c)^{2}+m a b c+n$.\n\nThen the function achieves it’s minimum value when $(a-b)(b-c)(c-a)=0$ or when $a b c=0$.\n\nIf $(a-b)(b-c)(c-a)=0$, then without loss of generality we may assume that $a=c$, and the given inequality is equivalent to\n$\n\\begin{aligned}\n& \\left(a^{2}+2 a b\\right)\\left(\\frac{1}{4 a^{2}}+\\frac{2}{(a+b)^{2}}\\right) \\geq \\frac{9}{4} \\\\\n& \\quad \\Leftrightarrow \\quad(a-b)^{2}\\left(\\frac{2 a+b}{2 a(a+b)^{2}}-\\frac{1}{(a+b)^{2}}\\right) \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad b(a-b)^{2} \\geq 0\n\\end{aligned}\n$\nas required.\nIf $a b c=0$, we may assume that $c=0$ and the given inequality becomes\n$\n\\begin{aligned}\n& a b\\left(\\frac{1}{(a+b)^{2}}+\\frac{1}{a^{2}}+\\frac{1}{b^{2}}\\right) \\geq \\frac{9}{4} \\quad \\Leftrightarrow \\quad(a-b)^{2}\\left(\\frac{1}{a b}-\\frac{1}{4(a+b)^{2}}\\right) \\geq 0 \\\\\n& \\quad \\Leftrightarrow \\quad(a-b)^{2}\\left(4 a^{2}+4 b^{2}+7 a b\\right) \\geq 0\n\\end{aligned}\n$\nand the problem is solved.\n\nEquality in the above analysis holds when two variables are equal and the third is zero, i.e., when $(a, b, c)$ is a permutation of $(t, t, 0)$ for $t > 0$. In this case, the minimum value of the expression is achieved, so $C = \frac{9}{4}$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{9}{4}$.

P243. Let $k \in \mathbb{N}$, and $a_{1}, a_{2}, \ldots, a_{n}$ be positive real numbers such that $a_{1} + a_{2} + \cdots + a_{n} = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a_{1}, a_{2}, \ldots, a_{n}$: $a_{1}^{-k} + a_{2}^{-k} + \cdots + a_{n}^{-k} \geq C.$

S243. $C = n^{k+1}$

Since $A M \geq G M$ we have\n$\n\\sqrt[n]{a_{1} a_{2} \\cdots a_{n}} \\leq \\frac{a_{1}+a_{2}+\\cdots+a_{n}}{n}=\\frac{1}{n}\n$\nor\n$\nn \\leq \\sqrt[n]{\\frac{1}{a_{1}} \\frac{1}{a_{2}} \\cdots \\frac{1}{a_{n}}}\n$\n\nHence\n$\nn^{k} \\leq \\sqrt[n]{a_{1}^{-k} a_{2}^{-k} \\cdots a_{n}^{-k}} \\leq \\frac{a_{1}^{-k}+a_{2}^{-k}+\\cdots+a_{n}^{-k}}{n}\n$\ni.e.\n$\na_{1}^{-k}+a_{2}^{-k}+\\cdots+a_{n}^{-k} \\geq n^{k+1}\n$\nas required.\n\nEquality holds when $a_1 = a_2 = \cdots = a_n = \frac{1}{n}$, since then $a_i^{-k} = n^k$ for each $i$, so the sum is $n \cdot n^k = n^{k+1}$. This gives the minimum value of the sum, so $C = n^{k+1}$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = n^{k+1}$.

P244. Let $a, b, c, d > 0$ such that $a+b+c=1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c, d$: $a^{3}+b^{3}+c^{3}+a b c d \geq C.$

S244. $C = \min \left{\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right}$

Suppose the inequality is false. Then we have, taking into account that $a b c \leq \frac{1}{27}$, the inequality $d\left(\frac{1}{27}-a b c\right)>a^{3}+b^{3}+c^{3}-\frac{1}{9}$. We may assume that $a b c<\frac{1}{27}$. Now, we will reach a contradiction proving that $a^{3}+b^{3}+c^{3}+a b c d \geq \frac{1}{4}$. It is sufficient to prove that\n\n$\n\\frac{a^{3}+b^{3}+c^{3}-\\frac{1}{9}}{\\frac{1}{27}-a b c} a b c+a^{3}+b^{3}+c^{3} \\geq \\frac{1}{4}\n$\n\nBut this inequality is equivalent to $4 \sum a^{3}+15 a b c \geq 1$. We use now the identity $\sum a^{3}=3 a b c+1-3 \sum a b$ and reduce the problem to proving that $\sum a b \leq \frac{1+9 a b c}{4}$, which is Schur’s Inequality.\n\nEquality in the above inequalities holds when $a = b = c = \frac{1}{3}$ and $d$ is arbitrary, since then $a^3 + b^3 + c^3 + a b c d = 3 \left(\frac{1}{3}\right)^3 + \left(\frac{1}{27}\right) d = \frac{1}{9} + \frac{d}{27}$. The minimum value of $a^3 + b^3 + c^3 + a b c d$ is also achieved when one of $a, b, c$ is $1$ and the others are $0$, giving $1^3 = 1$. However, since $a, b, c > 0$ and $a + b + c = 1$, the minimum is $\frac{1}{4}$ when two variables approach $0$ and one approaches $\frac{1}{2}$, i.e., $a = b = \frac{1}{2}, c \to 0$. Thus, the minimum value of $C$ is $\min \left\{\frac{1}{4}, \frac{1}{9} + \frac{d}{27}\right\}$.\n\nTherefore, the answer is $C = \min \left\{\frac{1}{4}, \frac{1}{9}+\frac{d}{27}\right\}$.

P245. Let $x, y, z > 0$ be real numbers such that $x + y + z = 1$. Determine the minimal constant $C$ such that the following inequality holds for all $x, y, z$: $\left(x^{2}+y^{2}\right)\left(y^{2}+z^{2}\right)\left(z^{2}+x^{2}\right) \leq C.$

S245. $C = \frac{1}{32}$

Let $p=x+y+z=1, q=x y+y z+z x, r=x y z$.\nThen we have\n$\n\\begin{aligned}\nx^{2}+y^{2} & =(x+y)^{2}-2 x y=(1-z)^{2}-2 x y=1-2 z+z^{2}-2 x y \\\\\n& =1-z-z(1-z)-2 x y=1-z-z(x+y)-2 x y=1-z-q-x y\n\\end{aligned}\n$\n\nAnalogously we deduce\n$\ny^{2}+z^{2}=1-x-q-y z \\quad \\text { and } \\quad z^{2}+x^{2}=1-y-q-z x\n$\n\nSo the given inequality becomes\n$\n\\begin{equation*}\n(1-z-q-x y)(1-x-q-y z)(1-y-q-z x) \\leq \\frac{1}{32} \\tag{1}\n\\end{equation*}\n$\n\nAfter algebraic transformations we find that inequality (1) is equivalent to\n$\n\\begin{equation*}\nq^{2}-2 q^{3}-r(2+r-4 q) \\leq \\frac{1}{32} \\tag{2}\n\\end{equation*}\n$\n\nAssume that $q \leq \frac{1}{4}$.\nUsing $ p^{3}-4 p q+9 r \geq 0$ , it follows that\n$\n9 r \\geq 4 q-1, \\quad \\text { i.e. } \\quad r \\geq \\frac{4 q-1}{9}\n$\nand clearly $q \leq \frac{1}{3}$.\nIt follows that\n$\n2+r-4 q \\geq 2+\\frac{4 q-1}{9}-4 q=\\frac{17-32 q}{9} \\geq \\frac{17-\\frac{32}{3}}{9}>0\n$\n\nSo we have\n$\n\\begin{aligned}\nq^{2}-2 q^{3}-r(2+r-4 q) & \\leq q^{2}-2 q^{3}=q^{2}(1-2 q) \\\\\n& =\\frac{q}{2} \\cdot 2 q(1-2 q) \\leq \\frac{q}{2}\\left(\\frac{2 q+(1-2 q)}{2}\\right)^{2}=\\frac{q}{8} \\leq \\frac{1}{32}\n\\end{aligned}\n$\ni.e. inequality (2) holds for $q \leq \frac{1}{4}$.\n\nWe need just to consider the case when $q>\frac{1}{4}$.\nLet\n$\n\\begin{equation*}\nf(r)=q^{2}-2 q^{3}-r(2+r-4 q) \\tag{3}\n\\end{equation*}\n$\n\nClearly $r \geq \frac{4 q-1}{9}$.\nUsing $p q-9 r \geq 0$ it follows that $9 r \leq q$, i.e. $r \leq \frac{q}{9}$.\nWe have\n$\nf^{\\prime}(r)=4 q-2-2 r \\leq \\frac{4}{3}-2-2 r \\leq 0\n$\n\nThis means that $f$ is a strictly decreasing function on $\left(\frac{4 q-1}{9}, \frac{q}{9}\right)$, from which it follows that\n$\nf(r) \\leq f\\left(\\frac{4 q-1}{9}\\right)=q^{2}-2 q^{3}-\\frac{1}{81}(4 q-1)(17-32 q)\n$\ni.e.\n$\n\\begin{equation*}\nf(r) \\leq \\frac{81\\left(q^{2}-2 q^{3}\\right)-(4 q-1)(17-32 q)}{81} \\tag{4}\n\\end{equation*}\n$\n\nLet\n$\n\\begin{equation*}\ng(q)=81\\left(q^{2}-2 q^{3}\\right)-(4 q-1)(17-32 q) \\tag{5}\n\\end{equation*}\n$\n\nThen\n$\ng^{\\prime}(q)=-486 q^{2}+418 q-100\n$\n\nSince $\frac{1}{4}<q \leq \frac{1}{3}$, we get\n$\ng^{\\prime}(q)=-486 q^{2}+418 q-100<\\frac{-486}{16}+\\frac{418}{3}-100<0\n$\n\nSo $g$ decreases on $(1 / 4,1 / 3)$, i.e. we have\n$\n\\begin{equation*}\ng(q)<g\\left(\\frac{1}{4}\\right)=\\frac{81}{32} \\tag{6}\n\\end{equation*}\n$\n\nFinally by (3), (4), (5) and (6) we obtain\n$\n\\begin{aligned}\nq^{2}-2 q^{3}-r(2+r-4 q) & =f(r) \\leq f\\left(\\frac{4 q-1}{9}\\right) \\\\\n& =\\frac{81\\left(q^{2}-2 q^{3}\\right)-(4 q-1)(17-32 q)}{81} \\\\\n& =\\frac{g(q)}{81}<\\frac{g\\left(\\frac{1}{4}\\right)}{81}=\\frac{\\frac{81}{32}}{81}=\\frac{1}{32}\n\\end{aligned}\n$\nas required.\n\nEquality holds when $x = y = z = \frac{1}{3}$, since then $x + y + z = 1$ and $\left(x^2 + y^2\right)\left(y^2 + z^2\right)\left(z^2 + x^2\right) = \left(2\left(\frac{1}{3}\right)^2\right)^3 = \left(\frac{2}{9}\right)^3 = \frac{8}{729} = \frac{1}{32}$. This gives the maximum value of the expression, so $C = \frac{1}{32}$ is minimal such that the inequality always holds.\n\nTherefore, the answer is $C = \frac{1}{32}$.

P246. Let $a, b, c$ be real numbers such that $a^3 + b^3 + c^3 - 3abc = 1$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: $a^2 + b^2 + c^2 \geq C$

S246. $C = 1$

Observe that\n$\n\\begin{aligned}\n1 & =a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right) \\\\\n& =\\frac{(a+b+c)}{2}\\left((a-b)^{2}+(b-c)^{2}+(c-a)^{2}\\right)\n\\end{aligned}\n$\n\nSince $(a-b)^{2}+(b-c)^{2}+(c-a)^{2} \geq 0$ we must have $a+b+c>0$.\nAccording to\n$\n(a+b+c)\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)=1\n$\nwe deduce\n$\n(a+b+c)\\left(a^{2}+b^{2}+c^{2}-\\frac{(a+b+c)^{2}-a^{2}-b^{2}-c^{2}}{2}\\right)=1\n$\nand easily find\n$\na^{2}+b^{2}+c^{2}=\\frac{1}{3}\\left((a+b+c)^{2}+\\frac{2}{a+b+c}\\right)\n$\n\nSince $a+b+c>0$ we may use $A M \geq G M$ as follows\n$\na^{2}+b^{2}+c^{2}=\\frac{1}{3}\\left((a+b+c)^{2}+\\frac{1}{a+b+c}+\\frac{1}{a+b+c}\\right) \\geq 1\n$\nas required.\nEquality occurs iff $a+b+c=1$.\n\nEquality holds when $a+b+c=1$, which is possible for example when $a = b = c = \frac{1}{3}$. In this case, $a^2 + b^2 + c^2 = 3 \times \left(\frac{1}{3}\right)^2 = 1$. Thus, the minimum value of $a^2 + b^2 + c^2$ is $1$, so $C = 1$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

P247. Let $a, b, c$ be positive real numbers such that $a + b + c = 1$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c$: $a b c + \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \geq C.$

S247. $C = \frac{244}{27}$

By the inequality $A M \geq G M$ we get\n$\na b c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq 4 \\sqrt[4]{a b c \\cdot \\frac{1}{a} \\cdot \\frac{1}{b} \\cdot \\frac{1}{c}}=4\n$\nwith equality if and only if $a b c=\frac{1}{a}=\frac{1}{b}=\frac{1}{c}$, from which we easily deduce that $a=b=c=1$ and then $a+b+c=3$, a contradiction since $a+b+c=1$.\n\nSince $a b c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ is symmetrical with respect to $a, b$ and $c$ we estimate that the minimal value occurs when $a=b=c$, i.e. $a=b=c=1 / 3$, since $a+b+c=1$.\n\nLet $a b c=\frac{1}{\alpha a}=\frac{1}{\alpha b}=\frac{1}{\alpha c}$, from which we obtain $\alpha=\frac{1}{a^{2} b c}=81$.\n\nTherefore let us rewrite the given expression as follows\n$\n\\begin{equation*}\na b c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}=a b c+\\frac{1}{81 a}+\\frac{1}{81 b}+\\frac{1}{81 c}+\\frac{80}{81}\\left(\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}\\right) \\tag{1}\n\\end{equation*}\n$\n\nBy $A M \geq G M$ and $A M \geq H M$ we have\n$\n\\begin{equation*}\na b c+\\frac{1}{81 a}+\\frac{1}{81 b}+\\frac{1}{81 c} \\geq 4 \\sqrt[4]{a b c \\cdot \\frac{1}{81 a} \\cdot \\frac{1}{81 b} \\cdot \\frac{1}{81 c}}=\\frac{4}{27} \\tag{2}\n\\end{equation*}\n$\nand\n$\n\\begin{equation*}\n\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq \\frac{9}{a+b+c}=9 \\tag{3}\n\\end{equation*}\n$\n\nBy (1), (2) and (3) we have\n$\na b c+\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c} \\geq \\frac{4}{27}+\\frac{80}{9}=\\frac{244}{27}\n$\nwith equality if and only if $a=b=c=\frac{1}{3}$.\n\nEquality is achieved when $a = b = c = \frac{1}{3}$, which satisfies $a + b + c = 1$. This gives the minimum value of the expression, so $C = \frac{244}{27}$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = \frac{244}{27}$.

P248. Let $a, b, c \in \mathbb{R}^{+}$ such that $a + 2b + 3c \geq 20$. Find the largest constant $C$ such that the following inequality holds for all $a, b, c$ satisfying the given constraint: $S = a + b + c + \frac{3}{a} + \frac{9}{2b} + \frac{4}{c} \geq C$

S248. $C = 13$

$S=13$ at the point $a=2, b=3, c=4$.\nUsing $A M \geq G M$ we get\n$\na+\\frac{4}{a} \\geq 2 \\sqrt{a \\cdot \\frac{4}{a}}=4, \\quad b+\\frac{9}{b} \\geq 2 \\sqrt{b \\cdot \\frac{9}{b}}=6, \\quad c+\\frac{16}{c} \\geq 2 \\sqrt{c \\cdot \\frac{16}{c}}=8\n$\ni.e.\n$\n\\frac{3}{4}\\left(a+\\frac{4}{a}\\right) \\geq 3, \\quad \\frac{1}{2}\\left(b+\\frac{9}{b}\\right) \\geq 3 \\quad \\text { and } \\quad \\frac{1}{4}\\left(c+\\frac{16}{c}\\right) \\geq 2\n$\n\nAdding the last three inequalities we have\n$\n\\begin{equation*}\n\\frac{3}{4} a+\\frac{1}{2} b+\\frac{1}{4} c+\\frac{3}{a}+\\frac{9}{2 b}+\\frac{4}{c} \\geq 8 \\tag{1}\n\\end{equation*}\n$\n\nUsing $a+2 b+3 c \geq 20$ we obtain\n$\n\\begin{equation*}\n\\frac{1}{4} a+\\frac{1}{2} b+\\frac{3}{4} c \\geq 5 \\tag{2}\n\\end{equation*}\n$\n\nFinally, after adding (1) and (2) we get\n$\na+b+c+\\frac{3}{a}+\\frac{9}{2 b}+\\frac{4}{c} \\geq 13\n$\nas desired.\n\nEquality holds when $a=2$, $b=3$, $c=4$, since at this point $a+2b+3c=2+6+12=20$ and $S=2+3+4+\frac{3}{2}+\frac{9}{6}+1=9+1.5+1.5+1=13$. This gives the minimum value of $S$, so $C=13$ is the largest constant for which the inequality always holds.\n\nTherefore, the answer is $C = 13$.

P249. Let $a, b, c, d > 0$ be real numbers. Let $u = ab + ac + ad + bc + bd + cd$ and $v = abc + abd + acd + bcd$. Find the maximal constant $C$ such that the following inequality holds for all $a, b, c, d > 0$: $2u^3 \geq Cv^2$

S249. $C = 27$

We have $p_{2}=\frac{u}{\binom{4}{2}}=\frac{u}{6}$ and $p_{3}=\frac{v}{\binom{4}{3}}=\frac{v}{4}$.\nBy Maclaurin’s inequality we have\n$\np_{2}^{\\frac{1}{2}} \\geq p_{3}^{\\frac{1}{3}} \\quad \\Leftrightarrow \\quad p_{2}^{3} \\geq p_{3}^{2} \\quad \\Leftrightarrow \\quad\\left(\\frac{u}{6}\\right)^{3} \\geq\\left(\\frac{v}{4}\\right)^{2} \\quad \\Leftrightarrow \\quad 2 u^{3} \\geq 27 v^{2}\n$\n\nEquality holds when $a = b = c = d$, that is, all variables are equal. In this case, the inequality becomes an equality, and $C = 27$ is the maximal value for which the inequality always holds.\n\nTherefore, the answer is $C = 27$.

P250. Let $x, y, z \in \mathbb{R}^{+}$ such that $x + y + z = 1$. Find the maximal constant $C$ such that the following inequality holds for all $x, y, z$ satisfying the given constraint: $\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} \geq C$

S250. $C = 1$

We have\n$\n\\begin{equation*}\n\\frac{x y}{z}+\\frac{y z}{x}+\\frac{z x}{y}=\\frac{1}{2}\\left(\\frac{x y}{z}+\\frac{y z}{x}\\right)+\\frac{1}{2}\\left(\\frac{y z}{x}+\\frac{z x}{y}\\right)+\\frac{1}{2}\\left(\\frac{z x}{y}+\\frac{x y}{z}\\right) \\tag{1}\n\\end{equation*}\n$\n\nSince $A M \geq G M$ we have\n$\n\\frac{1}{2}\\left(\\frac{x y}{z}+\\frac{y z}{x}\\right) \\geq \\sqrt{\\frac{x y}{z} \\frac{y z}{x}}=y\n$\n\nAnalogously we get\n$\n\\frac{1}{2}\\left(\\frac{y z}{x}+\\frac{z x}{y}\\right) \\geq z \\quad \\text { and } \\quad \\frac{1}{2}\\left(\\frac{z x}{y}+\\frac{x y}{z}\\right) \\geq x\n$\n\nAdding these three inequalities we obtain\n$\n\\frac{x y}{z}+\\frac{y z}{x}+\\frac{z x}{y} \\geq x+y+z=1\n$\n\nEquality holds if and only if $x = y = z$, that is, when $x = y = z = \frac{1}{3}$. In this case, the minimum value of the expression is achieved, so $C = 1$ is the maximal constant for which the inequality always holds.\n\nTherefore, the answer is $C = 1$.

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Inequality Proof Problems [201-250]

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